Month: April 2023

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory:

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Solution:

752 + 20f = 792 + 18.1
20f – 18f = 792 – 752
2f = 40
f = \(\frac{40}{2}\) = 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes:

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.
Solution:

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO2 (in ppm)	Frequency
0.00 – 0.04	4
0.04 – 0.08	9
0.08 – 0.12	9
0.12 – 0.16	2
0.16 – 0.20	4
0.20 – 0.24	2

Find the mean concentration of SO₂ in the air.
Solution:

Mean concentration of SO₂ in air = 0.099 ppm

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:


Mean number of days a student was absent is 12.48.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:

Jharkhand Board JAC Class 10 Maths Solutions Chapter 10 Circles Ex 10.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 10 Circles Exercise 10.1

Question 1.
How many tangents can a circle have?
Solution :
A circle can have infinite number of tangents.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ________ point (s).
(ii) A line intersecting a circle in two points is called a _______.
(iii) A circle can have _________ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ________.
Solution :
(i) one,
(ii) sccant,
(iii) two,
(iv) point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:

(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm.
Solution :
PQ² = OQ² – OP² (Using Pythagoras theorem)
= (12)² – (5)²
= 144 – 25
= 119.
PQ = \(\sqrt{119}\) cm.

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution :
AB is the given line. CD is the secant and PQ is the tangent to the circle at point R.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 9 Some Applications of Trigonometry Exercise 9.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
Solution :

sin \(\hat{\mathrm{C}}\) = \(\frac{AB}{AC}\)
sin 30° = \(\frac{1}{2}\)
\(\frac{1}{2}\) = \(\frac{AB}{20}\)
AB = \(\frac{20}{2}\) = 10 m

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution :

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:

(i) Figure (a) shows the slide for children below the age of 5 years.
Let BC = 1.5 m be the height of the slide. Slide AC is inclined at CAB = 30° to the ground.
In right angled ΔABC, sin 30° = \(\frac{BC}{AC}\)
\(\frac{1}{2}=\frac{15}{AC}\)
⇒ AC = 3 m.

(ii) Figure (b) shows the slide for elder children. Let RQ = 3 m be the height of the slide. Slide PR is inclined at ∠RPQ = 60° to the ground.
In right angled ΔPQR, sin 60° = \(\frac{RQ}{PR}\) ⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{3}{PR}\)
PR = \(\frac{3 \times 2}{\sqrt{3}}\) = 2\(\sqrt{3}\) m.

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution :

tan C = \(\frac{AB}{CB}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
tan 30° = \(\frac{h}{30}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{h}{30}\)
h\(\sqrt{3}\) = 30
h = \(\frac{30}{\sqrt{3}}=\frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{30 \sqrt{3}}{3}\)
h = 10\(\sqrt{3}\)mts.
Hence, height of the tower is 10\(\sqrt{3}\) mts.

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution :

Length of the string is 40\(\sqrt{3}\) mts.

Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution :

Let the boy be standing at point B initially. He walks towards the building and reaches point D. From the figure, the distance walked by the boy towards the building is BD.
AC = AG – CG
AC = 30 – 1.5 = 28.5
Now, in ΔABC, we have
tan 30 = \(\frac{AC}{BC}\)
BC = \(\frac{AC}{tan 30}\)
BC = 28.5\(\sqrt{3}\)
Again, in ΔADC, we have
tan 60 = \(\frac{AC}{DC}\)
DC = \(\frac{AC}{tan 60}\)
DC = \(\frac{28.5}{\sqrt{3}}\)
DC = \(\frac{28.5 \sqrt{3}}{3}\) = 9.5\(\sqrt{3}\)
BD = BC – DC
BD = 28.5\(\sqrt{3}\) – 9.5\(\sqrt{3}\)
BD = 19\(\sqrt{3}\)
The distance walked by the boy towards the building is 19\(\sqrt{3}\) m.

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution :

tan 60° = \(\sqrt{3}\)
tan 45° = 1
In ΔADC, A\(\hat{\mathrm{D}}\)C = 60
tan 60° = \(\frac{AC}{AD}\)
\(\sqrt{3}\) = \(\frac{x+20}{DA}\)

In ΔBDA, A\(\hat{\mathrm{D}}\)B = 45°
tan 45° = \(\frac{AB}{AD}\) = 1
AB = AD = 20 mts.
DA\(\sqrt{3}\) = x + 20
\(\sqrt{3}\)DA = x + 20
\(\sqrt{3}\)(20) = x + 20
x = 20\(\sqrt{3}\) – 20
= 20(\(\sqrt{3}\) – 1)
Height of the tower B = 20(\(\sqrt{3}\) – 1) mts.

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution :

tan 60° = \(\sqrt{3}\)
In ΔADC, tan 60° = \(\frac{AC}{CD}\)
\(\sqrt{3}\) = \(\frac{AC}{CD}\)
AC = \(\sqrt{3}\)CD

In ΔBDC, tan B\(\hat{\mathrm{D}}\)C = tan 45° = \(\frac{BC}{CD}\)
1 = \(\frac{BC}{CD}\)
CD = BC.

A = AC – CB
= AC – CD (∵ CB = CD)
= \(\sqrt{3}\)CD – CD
1.6 = CD(\(\sqrt{3}\) – 1)

Height of the pedestal = 0.8(\(\sqrt{3}\) + 1) mts.

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution :

Let the height of the building CD be h.
tan 60° = \(\sqrt{3}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
In ΔABC, tan \(\hat{\mathrm{C}}\) = \(\frac{AB}{AC}\)
tan 60° = \(\frac{50}{AC}\)
\(\sqrt{3}\) = \(\frac{50}{AC}\)
AC\(\sqrt{3}\) = 50°
AC = \(\frac{50}{\sqrt{3}}\)mts.

Height of the building = 16\(\frac{2}{3}\) mts.

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution :

Let AE = x.
EB = 80 – x
tan 60° = \(\sqrt{3}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)

Question 11.
A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Solution :

tan 60° = \(\sqrt{3}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
In ΔABC, tan 60° = \(\frac{AB}{BC}\) = \(\frac{h}{x}\)
\(\frac{h}{x}\) = \(\sqrt{3}\)
∴ h = \(\sqrt{3}\)x
In ΔADB, tan 30° = \(\frac{AB}{DB}\)

Height = x\(\sqrt{3}\) = 10\(\sqrt{3}\) mts. Width of the canal = 10 mts.

Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution :

In ΔDBC, tan 60° = \(\frac{DC}{BC}\)
\(\sqrt{3}\) = \(\frac{DC}{BC}\)
= \(\frac{DC}{AE}\) = \(\frac{DC}{7}\) (BC = AE).

In ΔABE, tan 45° = \(\frac{AB}{AE}\)
1 = \(\frac{7}{AE}\)
∴ AE = 7 mts.
AB = BC = 7 mts.
DC = 7\(\sqrt{3}\)
∴ DE = DC + CE
= 7\(\sqrt{3}\) + 7 = 7(\(\sqrt{3}\) + 1) mts.

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution :

Hence, the distance between the two ships is 75(\(\sqrt{3}\) – 1) m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Solution :

Let the initial position of the balloon be A and final position be B.
Height of the balloon above the girl’s height = 88.2 m – 1.2 m = 87m
Distance travelled by the balloon = DE = CE – CD

Distance travelled by the balloon, DE = CE – CD
= (87\(\sqrt{3}\) – 29\(\sqrt{3}\)) m
= 29\(\sqrt{3}\) (3 – 1) m
= 58\(\sqrt{3}\) m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution :

In ΔBCA, tan 30° = \(\frac{h}{CA}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{h}{CA}\)
CA = h\(\sqrt{3}\) mts.

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution :

Let AB be the tower. ∠ABC = x
∴ ∠ADB = 90° – x
In ΔABC tan x = \(\frac{AB}{BC}\)
tan x = \(\frac{AB}{4}\) ……………(i)
In ΔADB tan (90° – x) = \(\frac{AB}{9}\)
cot x = \(\frac{AB}{9}\) ……………(ii)
(i) × (ii)
tan x × cot x = \(\frac{AB}{4}\) × \(\frac{AB}{9}\)
tan x × \(\frac{1}{tan x}\) = \(\frac{\mathrm{AB}^2}{36}\)
1 = \(\frac{\mathrm{AB}^2}{36}\)
AB² = 36
AB = ± \(\sqrt{36}\)
AB = ± 6
∴ Height of the tower AB = 6 m.
Note: C and D can be taken on the same side of AB.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution :
(i) We know that, cosec²A = 1 + cot² A ⇒ cosec A = \(\sqrt{1+\cot ^2 \mathrm{~A}}\)

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution :
(i) sin A sin² A + cos² A = 1
sin² A = 1 – cos² A

Question 3.
Evaluate:
(i) \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)
(ii) sin 25° cos 65° + cos 25° sin 65°.
Solution :
(i) \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)
sin (90 – θ) = cos θ
sin (90 – 27°) = cos 27°
sin 63° = cos 27°
sin² 63 = cos² 27°

cos (90 – θ) = sin θ
cos (90° – 73°) = sin 73°
cos (17°) = sin 73°
cos² 17° = sin² 73°

(ii) sin 25° cos 65° + cos 25° sin 65°
sin (90° – θ) = cos θ
sin (90° – 25°) = cos 25°
sin (90° – 65°) = cos 65°
sin 25° = cos 65°

cos (90° – θ) = sin θ
cos (90° – 65°) = sin 65°
cos 25° = sin 65°

sin 25° = cos 65°
cos 65° . cos 65° + sin 65° . sin 65°
∴ cos² 65 + sin² 65 = 1

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan² A=
(A) 1
(B) 9
(C) 8
(D) 0
Solution :
9 sec² A – 9 tan² A
9(1 + tan² A) – 9 tan² A
9 + 9 tan² A – 9 tan² A = 9.
∴ 9 sec² A – 9 tan² A = 9.

(ii) (1 + tan θ + sec θ) (1 + cot 0 – cosec θ) =
(A) 0
(B) 1
(C) 2
(D) – 1
Solution :

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A
Solution :

(iv) \(\frac{1+\tan ^2 A}{1+\cot ^2 A}\) =
(A) sec² A
(B) – 1
(C) cot² A
(D) tan² A
Solution :

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
Solution :

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

LHS = sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A [∵ (a + b)² = a² + b² + 2ab]
= (sin² A + cos² A) + (1 + cot² A) + 2 sin A . \(\frac{1}{sin A}\) + (1 + tan² A) + 2 cos A . \(\frac{1}{cos A}\) (∵ 1 + cot² A = cosec² A and sec² A = 1 + tan² A)
= 1 + 1 + cot² A + 2 + 1 + tan² A + 2 (∵ sin² A + cos² A = 1)
= 7 + tan² A + cot² A = RHS.

(ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
Solution :
[Hint: Simplify LHS and RHS separately].
LHS = (cosec A – sin A) (sec A – cos A)

Jharkhand Board JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 15 Probability Exercise 15.1

Question 1.
Complete the following statements:
i) Probability of an event E+ Probability of the event ‘not E’ =
ii) The probability of an event that cannot happen is …………….. Such an event is called ……………
iii) The probability of an event that is certain to happen is …………. Such an event is called …………….
iv) The sum of the probabilities of all the elementary events of an experiment is ……………
v) The probability of an event is greater than or equal to ………………. and less than or equal to ………………
Solution:
i) 1,
ii) 0, impossible event,
iii) 1, sure or certain,
iv) 1,
v) 0, 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
i) A driver attempts to start a car. The car starts or does not start.
ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
iii) A trial made to answer a true-false question. The answer is right or wrong.
iv) A baby is born. It is a boy or a girl.
Solution:
(iii) and (iv) have equally likely outcomes. Only two possibilities are there in each of these cases.

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
When a coin is tossed, head or tail are equally likely possible events. So the result of an individual coin toss is unpredictable.

Question 4.
Which of the following cannot be the probability of an event:
A) \(\frac{2}{3}\)
B) -1.5
C) 15%
D) 0.7?
Solution:
(B) Because, probability of an event cannot be negative.

Question 5.
If P(E)= 0.05, what is the probability of ‘not E’?
Solution:
P(E) = 0.05.
[P(\(\overline{\mathrm{E}}\)) = Probability of not an event]
We know that P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(E) = 1 – 0.05 = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
i) an orange flavoured candy?
ii) a lemon flavoured candy?
Solution:
i) P(orange flavoured candy) = 0. Impossible event.
ii) P(Lemon flavoured candy) = 1. Sure event.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let E = Event of 2 students not having the same birthday
∴ P(E) = 0.992
∴ P(E) + P(\(\overline{\mathrm{E}}\)) = 1
∴ 0.992 + P(\(\overline{\mathrm{E}}\)) = 1
⇒ P(\(\overline{\mathrm{E}}\)) = 1 – 0.992
= 0.008
So, the probability of two students having the same birthday is 0.008.

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Solution:
Total number of balls, n(S) = 3 + 5 = 8.
Let E = Event of drawing 1 red ball
∴ n(E) = 3
(i) Probability of drawing a red ball = \(\frac{n(E)}{n(S)}=\frac{3}{8}\)
(ii) Probability of not drawing a red ball = 1 – P(Drawing a red ball)
= \(1-\frac{3}{8}=\frac{5}{8}\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Total possible outcomes = 5 + 8 + 4 = 17.
P(R) = \(\frac{5}{17}\), P(W) = \(\frac{8}{17}\)
P(Not green) = P(R + W) = \(\frac{5}{17}+\frac{8}{17}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50 p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a Rs. 5 coin?
Solution:
Total possible outcomes: 100 + 50 + 20 + 10 = 180.
P(50 paise coin) = \(\frac{100}{180}=\frac{5}{9}\)
P(Not Rs. 5 coin) = \(\frac{100}{180}+\frac{50}{180}+\frac{20}{180}\)
= \(\frac{170}{180}=\frac{17}{18}\).

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Solution:
Total number of fish in an aquarium = 5 male fish + 8 female fish = 13 fish
∴ Probability of taking out a male fish = \(\frac{\text { Number of male fish }}{\text { Total number of fish }}=\frac{5}{13}\)

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at
i) 8?
ii) an odd number?
iii) a number greater than 2?
iv) a number less than 9?

Solution:
Total possible outcomes = 8.
i) P(8) = \(\frac{1}{8}\)
ii) P(odd number) = \(\frac{4}{8}=\frac{1}{2}\)
iii) P(no. > 2) = \(\frac{6}{8}=\frac{3}{4}\)
iv) P(no. < 9) = \(\frac{8}{8}\) = 1

Question 13.
A die is thrown once. Find the probability of getting
(i) a prime number, (ii) a number lying between 2 and 6, (iii) an odd number.
Solution:
Total possible outcomes 1, 2, 3, 4, 5, 6 = 6
P(Prime number) (2, 3, 5) = \(\frac{3}{6}=\frac{1}{2}\)
P(Number between 2 and 6) = \(\frac{3}{6}=\frac{1}{2}\)
P(Odd number) = \(\frac{3}{6}=\frac{1}{2}\)

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
i) a king of red colour
ii) a face card
iii) a red face card
iv) the jack of hearts
v) a spade
vi) the queen of diamonds.
Solution:
Total number of cards in one deck, n(S) = 52.
i) Let E₁ = Event of getting a king of red colour
∴ n(E₁) = 2
(∵ In a deck of cards, 26 cards are red and 26 cards are black. There are four kings in a deck in which two are red and two are black)
Probability of getting a king of red colour = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{52}=\frac{1}{26}\)

ii) Let E₂ = Event of getting a face card
∴ n(E₂) = 12
(∵ In a deck of cards, there are 12 face cards – 4 king, 4 jack, 4 queen)
Probability of getting a face card = \(\frac{n\left(E_2\right)}{n(S)}=\frac{12}{52}=\frac{3}{13}\)

iii) Let E₃ = Event of getting a red face card
∴ n(E₃) = 6
(∵ In a deck of cards, there are 12 face cards – 6 red, 6 black)
Probability of getting a red face card = \(\frac{n\left(E_3\right)}{n(S)}=\frac{6}{52}=\frac{3}{26}\)

iv) Let E₄ = Event of getting a jack of hearts
∴ n(E₄) = 1
(∵ There are four jacks in a deck- 1 heart, 1 club, 1 spade, 1 diamond)
Probability of getting a jack of hearts = \(\frac{n\left(E_4\right)}{n(S)}=\frac{1}{52}\)

v) Let E₅ = Event of getting a spade
∴ n(E₅) = 13
(∵ In a deck of cards, there are 13 spades, 13 clubs, 13 hearts, 13 diamonds)
Probability of getting a spade = \(\frac{n\left(E_5\right)}{n(S)}=\frac{13}{52}=\frac{1}{4}\)

vi) Let E₆ = Event of getting a queen of diamond
∴ n(E₆) = 1
(∵ In 13 diamond cards, there is only one queen)
Probability of getting a queen of diamond = \(\frac{n\left(E_6\right)}{n(S)}=\frac{1}{52}\).

Question 15.
Five cards – the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random.
i) What is the probability that the card is the queen?
ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
i) Total possible outcomes = 5
P(Queen card) = \(\frac{1}{5}\)
ii) If the queen card is put aside, total possible outcomes = 4.
iii) P(ace) = \(\frac{1}{4}\)
iv) P(queen) = \(\frac{0}{2}\) = 0.

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Total possible outcomes = 132 + 12 = 144
No. of good pens = 132.
P(good pen) = \(\frac{132}{144}=\frac{11}{12}\)

Question 17.
i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
i) Total possible outcomes = 20
P(Defective bulbs) = \(\frac{4}{20}=\frac{1}{5}\)

ii) Total possible outcomes = 20 – 1 = 19
No. of defective bulbs = 4
No. of good bulbs = 15
P(Not defective bulb) = \(\frac{15}{19}\)

Question 18.
A box contains 90 dises which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number, (ii) a perfect square number, (iii) a number divisible by 5.
Solution:
S = {1, 2, 3, 4, 5, … 90}
∴ Total possible outcomes n(S) = 90
i) Number of 2-digit numbers = 90 – 9 = 81
P(a 2-digit number) = \(\frac{81}{90}=\frac{9}{10}\)

ii) Event A = {A perfect square number}
A = (1, 4, 9, 16, 25, 36, 49, 64, 81) = n(A) = 9
Probability of the event P(A) = \(\frac{n(A)}{n(S)}=\frac{9}{90}\)

iii) A number divisible by 5, i.e., multiples of 5.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90 = 18.
P(a no. divisible by 5) = \(\frac{18}{90}=\frac{1}{5}\).

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Total possible outcomes = 6
No. of A’s = 2
No. of D’s = 1
P(A) = \(\frac{2}{6}=\frac{1}{3}\)
P(D) = \(\frac{1}{6}\)

Question 20.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
i) she will buy it?
ii) she will not buy it?
Solution:
Total number of possible outcomes = 144
No. of good pens = 144 – 20 = 124.
P(of buying) = \(\frac{124}{144}=\frac{31}{36}\)
P(of not buying) = \(\frac{20}{144}=\frac{5}{36}\)

Question 21.
(i) Two dice, one blue and one grey, are thrown at the same time. Write down all possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8, (ii) 13, (iii) less than or equal to 12? Complete the following table:

Solution:


1) Sum of 2 dice = 2 (1 + 1)
P(Sum 2) = \(\frac{1}{36}\)

2) Sum of 2 dice = 3 (1 + 2) (2 + 1)
P(Sum 3) = \(\frac{2}{36}\)

3) Sum 4 (1, 3) (2, 2) (3, 1)
P(Sum 4) = \(\frac{3}{36}\)

4) Sum 5 (1, 4) (2, 3) (3, 2) (4, 1)
P(Sum 5) = \(\frac{4}{36}\)

5) Sum 6 (1, 5) (2, 4) (3, 3) (4, 2) (5, 1)
P(Sum 6) = \(\frac{5}{36}\)

6) Sum 7 (1,6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)
P(Sum 7) = \(\frac{6}{36}\)

7) Sum 8 (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)
P(Sum 8) = \(\frac{5}{36}\)

8) Sum 9 (3, 6) (4, 5) (5, 4) (6, 3)
P(Sum 9) = \(\frac{4}{36}\)

9) Sum 10 (4, 6) (5, 5) (6, 4)
P(Sum 10) = \(\frac{3}{36}\)

10) Sum 11 (5, 6) (6,5)
P(Sum 11) = \(\frac{2}{36}\)

11) Sum 12 (6,6)
P(Sum 12) = \(\frac{1}{36}\)

ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.
Solution:
Total possible outcomes of throwing the two dice, S =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6,3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36

a) Let E₁ = Sum of two dice is 3 = {(1, 2), (2, 1)}
n(E₁) = 2
∴ P(E₁) = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{36}\)

b) Let E₂ = Sum of two dice is 4 = {(1, 3), (2, 2), (3, 1)}
n(E₂) = 3
∴ P(E₂) = \(\frac{n\left(E_2\right)}{n(S)}=\frac{3}{36}\)

c) Let E₃ = Sum of two dice is 5 = {(1, 4), (2, 3), (3,2), (4, 1)}
n(E₃) = 4
∴ P(E₃) = \(\frac{n\left(E_3\right)}{n(S)}=\frac{4}{36}\)

d) Let E₄ = Sum of two dice is 6 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
n(E₄) = 5
∴ P(E₄) = \(\frac{5}{36}\)

e) Let E₅ = Sum of two dice is 7 = {(1, 6), (2, 5), (3, 4), (4,3), (5, 2), (6, 1)}
n(E₅) = 6
∴ P(E₅) = \(\frac{n\left(E_5\right)}{n(S)}=\frac{6}{36}\)

f) Let E₆ = Sum of two dice is 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6,2)}
n(E₆) = 5
∴ P(E₆) = \(\frac{n\left(E_6\right)}{n(S)}=\frac{5}{36}\)

g) Let E₇ = Sum of two dice is 9 = {(3, 6), (4, 5), (5, 4), (6,3)}
n(E₇) = 4
∴ P(E₇) = \(\frac{n\left(E_7\right)}{n(S)}=\frac{4}{36}\)

h) Let E₈ = Sum of two dice is 10 = {(4, 6), (5, 5), (6, 4)}
n(E₈) = 3
∴ P(E₈) = \(\frac{n\left(E_8\right)}{n(S)}=\frac{3}{36}\)

i) Let E₉ = Sum of two dice is 11 = {(6,5), (5, 6)}
n(E₉) = 2
∴ P(E₉) = \(\frac{n\left(E_9\right)}{n(S)}=\frac{2}{36}\)

j) Let E₁₀ = Sum of two dice is 12 = {(6, 6)}
n(E₁₀) = 1
∴ P(E₁₀) = \(\frac{n\left(E_{10}\right)}{n(S)}=\frac{1}{36}\)
No. The eleven events are not equally likely.

Question 22.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Total possible outcomes (H + T)³
= H³ + 3H²T + 3HT² + T³
HHH HHT HTH THH HTT THT TTH TTT = 8.
Possible losses HHT HTH THH HTT THT TTH = 6
P(of losses) = \(\frac{6}{8}=\frac{3}{4}\)

Question 23.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].
Solution:
i) Total number of cases, n(S) = 6² = 36
Let \(\overline{\mathrm{E}}\) = Event that 5 will come up either time
= {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
⇒ n(\(\overline{\mathrm{E}}\)) = 11
and E = Event that 5 will not come up either time
n(E) = 36 – 11 = 25
∴ Probability that 5 will not come up either time = \(1-\frac{11}{36}=\frac{36-11}{36}\)
= \(\frac{25}{36}\)
ii) Probability that 5 will come up at least once = 12 – 1 = \(\frac{1}{2}\).

Question 24.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Solution:
i) Incorrect: We can classify the outcomes like this but they are not then ‘equally likely’. The reason is that ‘one of each’ can result in two ways – from head on first coin and tail on the second coin or from tail on the first coin and head on the second coin. This makes it twice as likely as 2 heads or 2 tails.

ii) Correct. The two outcomes considered in the question are equally likely. Both have the same probability. i.e., \(\frac{1}{2}\).

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°.
Solution :
\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) sin θ = cos (90° – θ)
sin 18° = cos (90° – 18°) = cos 72°
∴ \(\frac{sin 18}{cos 72}\) = \(\frac{cos 72°}{cos 72°}\) = 1

(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\) tan θ = cot (90 – θ)
tan 26 = cot (90 – 26) = cot 64°.
∴ \(\frac{tan 26°}{cot 64°}\) = \(\frac{cot 64}{cot 64}\) = 1

(iii) cos 48° – sin 42° = cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0.

(iv) cosec 31° – sec 59° . cosec θ = sec (90 – θ)
cosec 31 = sec (90 – 31) = sec 59°.
cosec 31° – sec 59° = sec 59° – sec 59° = 0.

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1.
(ii) cos 38° cos 52° – sin 38° sin 52° = 0.
Solution :
(i) tan 48° tan 23° tan 42° tan 67° = 1.
LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° . tan 23° tan (90° – 48°) . tan (90° – 23°)
= tan 48° . tan 23° . cot 48° . cot 23° [∵ tan (90° – θ) = cot θ]
= tan 48° . tan 23° . \(\frac{1}{tan 48°}\) . \(\frac{1}{tan 23°}\) [∵ cot θ = \(\frac{1}{tan θ}\)]
= 1 = RHS.

(ii) cos 38° cos 52° – sin 38° sin 52° = 0.
LHS = cos 38° . cos 52° – sin 38° sin 52°
= cos (90° – 52°) cos (90° – 38°) – sin 38°. sin 52°
= sin 52° . sin 38° – sin 38° . sin 52° [∵ cos (90° – θ) = sin θ]
= 0 = RHS.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution :
tan θ = cot (90° – θ)
tan 2A = cot (90° – 2A)
tan 2A = cot (A – 18°) ∵ cot (90° – 2A) = cot (A – 18°)
90° – 2A = A – 18
– 2A – A = – 18 – 90
– 3A = – 108
A = \(\frac{-108}{-3}\) = 36°.

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution :
tan A = cot B (90° – A)
tan A = cot B
∴ cot (90° – A) = cot B
∴ 90° – A = B
90° = B + A
A + B = 90°.

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution :
sec 4A = cosec (A – 20°)
cosec (90° – 4A) = cosec (A – 20°)
90° – 4A = A – 20°
– 4A – A = – 110°
– 5A = – 110°
A = 22°.

Question 6.
If A, B and C are interior angles of a triangle ABC, then show that
Solution :
sin (\(\frac{B+C}{2}\)) = cos \(\frac{A}{2}\)
sin θ = cos (90° – θ)

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution :
sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°. (Since cos (90° – θ) = sin θ and sin (90° – θ) = cos θ)

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1

Question 1.
Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan² 45° + cos² 30° – sin² 60°

Solution :
(i) sin 60° cos 30° + sin 30° cos 60°
\(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\) + \(\frac{1}{2} \times \frac{1}{2}\)
\(\frac{3}{4}+\frac{1}{4}\) = 1

(ii) 2 tan² 45° + cos² 30° – sin² 60°
2(tan 45)² + (cos 30)² – (sin 60)²
2(1)² + (\(\frac{\sqrt{3}}{2}\))² – (\(\frac{\sqrt{3}}{2}\))² = 2

Question 2.
Choose the correct option and justify your choice:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°
Solution :

(ii) \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}\) =
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0
Solution :
\(\frac{1-(-1)^2}{1+(1)^2}=\frac{0}{2}\) = 0

(iii) sin 2A = 2 sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°
Solution :
sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 2(0) = 0.

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\) =
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution :

Question 3.
If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\), 0° < A + B ≤ 90°; A > B, find A and B.
Solution :
tan(A + B) = \(\sqrt{3}\)
tan 60° = \(\sqrt{3}\)
∴ A + B = 60°
tan (A – B) = \(\frac{1}{\sqrt{3}}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ A – B = 30°
∴ \(\hat{\mathbf{A}}\) = 45°, \(\hat{\mathbf{B}}\) = 15°

A + B = 60°
45 + B = 60°
B = 60 – 45 = 15
∴ B = 15°

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution :
i) Let A = 30°, B = 60°.
sin 30° = \(\frac{1}{2}\)
sin 30 + sin 60°
sin (A + B) = sin 90° = 1
sin 60° = \(\frac{\sqrt{3}}{2}\)
\(\frac{1}{2}+\frac{\sqrt{3}}{2}\) = \(\frac{1+\sqrt{3}}{2}\)
sin (A+B) sin A + sin B
False.

iv) [θ = 0] sin 0 = 0
cos 0 = 1

θ = 30° sin 30° = \(\frac{1}{2}\)
cos 30° = \(\frac{\sqrt{3}}{2}\)

θ = 45° sin 45° = \(\frac{1}{\sqrt{2}}\)
cos 45° = \(\frac{1}{\sqrt{2}}\)

θ = 60° sin 60° = \(\frac{\sqrt{3}}{2}\)
cos 60° = \(\frac{1}{2}\)

θ = 90° sin 90° = 1
cos 90° = 0
False, because it is true only for θ = 45°.

(v) cot A = \(\frac{cos A}{sin A}\) . cot 0° = \(\frac{cos 0°}{sin 0°}\) = \(\frac{1}{0}\) = Undefined. True.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.3

Question 1.
Find the sum of following APs:
1. 2, 7, 12, ……, to 10 terms.
2. -37, -33, -29, ……, to 12 terms.
3. 0.6, 1.7, 2.8, ……, to 100 terms.
4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots\) to 11 terms.
Solution:
1. For the given AP 2, 7, 12, ……., a = 2.
d = 7 – 2 = 5, n = 10 and S_n is to be found.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₁₀ = \(\frac{10}{2}\)[4 + (10 – 1) 5]
= 5(49) = 245
Thus, the sum of first 10 terms of the given AP is 245.

2. For the given AP -37, -33, -29, a = -37, d = (-33) – (-37) = 4, n = 12 and S_n is to be found.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₁₂ = \(\frac{12}{2}\)[-74 + (12 – 1)4]
= 6(-30) = – 180
Thus, the sum of first 12 terms of the given AP is -180.

3. For the given AP 0.6, 1.7, 2.8,…… a = 0.6, d = 1.7 – 0.6 = 1.1, n = 100 and S_n is to be found.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀₀ = \(\frac{100}{2}\)[1.2 + (100 – 1) (1.1)]
= 50[1.2 + 108.9]
= 50 × 110.1 = 5505
Thus, the sum of first 100 terms of the given AP is 5505.

4.

Thus, the sum of first 11 terms of the given AP is \(\frac{33}{20}\).

Question 2.
Find the sums given below:
1. 7 + 10\(\frac{1}{2}\) + 14 + … + 84
2. 34 + 32 + 30 + … + 10
3. (-5) + (-8) + (-11) + … + (-230)
Solution:
1. 7 + 10\(\frac{1}{2}\) + 14 + …… + 84
Here, a = 7; d = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\); last term l = 84.
Let the last term be nth term.
a_n = a + (n – 1)d
∴ 84 = 7 + (n – 1) (3\(\frac{1}{2}\))
∴ 77 = \(\frac{7}{2}\)(n – 1)
∴ (n – 1) = 22
∴ n = 23
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{23}{2}\)(7 + 84)
= \(\frac{23}{2}\) × 91 = 1046\(\frac{1}{2}\)
Thus, the required sum is 1046\(\frac{1}{2}\).

2. 34 + 32 + 30 + … + 10
Here, a = 34; d = 32 – 34 = (-2); last term l = 10.
Let the last term be nth term.
a_n = a + (n – 1)d
∴ 10 = 34 + (n – 1)(-2)
∴ -24 = -2(n – 1)
∴ (n – 1) = 12
∴ n = 13
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{13}{2}\)(34 + 10)
= 13 × 22 = 286
Thus, the required sum is 286.

3. (-5) + (-8) + (-11) + … + (-230)
Here, a = (-5); d = (-8) – (-5) = (-3):
last term l = (-230).
Let the last term be nth term.
a_n = a + (n – 1)d
∴ -230 = -5 + (n – 1)(-3)
∴ -225 = -3 (n – 1)
∴ n – 1 = 75
∴ n = 76
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{76}{2}\)[(-5) + (-230)]
= 38(-235) = -8930
Thus, the required sum is -8930.

Question 3.
In an AP:
1. Given a = 5, d = 3, a_n = 50, find n and S_n.
2. Given a = 7, a₁₃ = 35, find d and S₁₃.
3. Given a₁₂ = 37, d = 3, find a and S₁₂.
4. Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
5. Given d = 5, S₉ = 75, find a and a₉.
6. Given a = 2, d = 8, S_n = 90, find n and a_n.
7. Given a = 8, a_n = 62, S_n = 210, find n and d.
8. Given a_n = 4, d = 2, S_n = -14, find n and a.
9. Given a = 3, n = 8, S_n = 192, find d.
10. Given l = 28, S_n = 144, and there are total 9 terms. Find a.
Solution:
1. a = 5, d = 3, a_n = 50, n = ? S_n = ?
a_n = a + (n – 1)d
∴ 50 = 5 + (n – 1)3
∴ 45 = 3(n – 1)
∴ 15 = n – 1
∴ n = 16
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₆ = \(\frac{16}{2}\)(5 + 50)
∴ S₁₆ = 8 × 55
∴ S₁₆ = 440

2. a = 7, a₁₃ = 35, d = ?, S₁₃ = ?
a_n = a + (n – 1)d
a₁₃ = a + (13 – 1) d
∴ 35 = 7 + 12d
∴ 28 = 12d
∴ d = \(\frac{28}{12}\)
∴ d = \(\frac{7}{3}\)
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₃ = \(\frac{13}{2}\)(17 + 35)
∴ S₁₃ = 13 × 21
∴ S₁₃ = 273

3. a₁₂ = 37, d = 3, a = ?, S₁₂ = ?
a_n = a + (n – 1)d
∴ a₁₂ = a + 11d
∴ 37 = a + 11 (3)
∴ a = 4
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₂ = \(\frac{12}{2}\)(4 + 37)
∴ S₁₂ = 246

4. a₃ = 15, S₁₀ = 125, d = ?, a₁₀ = ?
a_n = a + (n – 1)d
∴ a₃ = a + 2d
∴ a + 2d = 15 …….(1)
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀ = \(\frac{10}{2}\)[2a + 9d]
∴ S₁₂₅ = 5(2a + 9d)
∴ 2a + 9d = 25 …….(2)
Solving equations (1) and (2), we get
d = -1 and a = 17.
a_n = a + (n – 1)d
∴ a₁₀ = a + 9d
∴ a₁₀ = 17 + 9(-1)
∴ a₁₀ = 8

5. d = 5, S₉ = 75, a = ?, a₉ = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₉ = \(\frac{9}{2}\)[2a + (9 – 1) d]
∴ 75 = \(\frac{9}{2}\)[2a + 8(5)]
∴ 75 = 9(a + 20)
∴ \(\frac{25}{3}\) = a + 20
∴ a = \(\frac{25}{3}\) – 20
∴ a = –\(\frac{35}{3}\)
a_n = a + (n – 1) d
∴ a₉ = a + 8d
∴ a₉ = (-\(\frac{35}{3}\)) + 8(5)
∴ a₉ = –\(\frac{35}{3}\) + 40
∴ a₉ = \(\frac{85}{3}\)

6. a = 2, d = 8, S_n = 90, n = ?, a_n = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 90 = \(\frac{n}{2}\)[4 + (n – 1)8]
∴ 90 = \(\frac{n}{2}\)[8n – 4]
∴ 90 = n (4n – 2)
∴ 4n² – 2n – 90 = 0
∴ 2n² – n – 45 = 0
∴ 2n² – 10n + 9n – 45 = 0
∴ 2n (n – 5) + 9 (n – 5) = 0
∴ (n – 5) (2n + 9) = 0
∴ n – 5 = 0 or 2n + 9 = 0
∴ n = 5 or n = –\(\frac{9}{2}\)
Since n is a positive integer, n ≠ –\(\frac{9}{2}\)
∴ n = 5
a_n = a + (n – 1)d
∴ a₅ = a + 4d
∴ a₅ = 2 + 4(8)
∴ a₅ = 34

7. a = 8, a_n = 62, S_n = 210, n = ? d = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ S_n = \(\frac{n}{2}\)(a + a_n)
∴ 210 = \(\frac{n}{2}\)(8 + 62)
∴ 420 = n (70)
∴ n = 6
a_n = a + (n – 1)d
∴ a₆ = a + 5d
∴ 62 = 8 + 5d
∴ 54 = 5d
∴ d = \(\frac{54}{5}\)

8. a_n = 4, d = 2, S_n = -14, n = ?, a = ?
a_n = a + (n – 1)d
∴ 4 = a + (n – 1) (2)
∴ 4 = a + 2n – 2
∴ a = 6 – 2n
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ -14 = \(\frac{n}{2}\)[2 (6 – 2n) + (n – 1) (2)] (by (1))
∴ -14 = \(\frac{n}{2}\)[12 – 4n + 2n – 2]
∴ -14 = \(\frac{n}{2}\)[-2n + 10]
∴ -14 = n(-n + 5)
∴ -14 = -n² + 5n
∴ n² – 5n – 14 = 0
∴ (n – 7)(n + 2) = 0
∴ n – 7 = 0 or n + 2 = 0
∴ n = 7 or n = -2
Since n is a positive integer, n ≠ -2.
∴ n = 7
By (1), a = 6 – 2n
∴ a = 6 – 2(7)
∴ a = -8

9. a = 3, n = 8, S_n = 192, d = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 192 = \(\frac{8}{2}\)[6 + (8 – 1) d]
∴ 192 = 4[6 + 7d]
∴ 48 = 6 + 7d
∴ 42 = 7d
∴ d = 6

10. l = 28, S_n = 144, n = 9, a = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ 144 = \(\frac{9}{2}\)(a + 28)
∴ 32 = (a + 28)
∴ a = 4

Question 4.
How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?
Solution:
Here, a = 9; d = 17 – 9 = 8; S_n = 636, n = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 636 = \(\frac{n}{2}\)[18 + (n – 1)8]
∴ 636 = \(\frac{n}{2}\)[10 + 8n]
∴ 636 = n[4n + 5]
∴ 4n² + 5n – 636 = 0
Here, a = 4; b = 5; c = -636
b² – 4ac = (5)² – 4(4)(-636)
= 25 + 10176
= 10201
∴ \(\sqrt{b^2-4 a c}=\sqrt{10201}=101\)
Then, n = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ n = \(\frac{-5 \pm 101}{8}\)
∴ n = \(\frac{96}{2}\) or n = \(\frac{-106}{8}\)
∴ n = 12 or n = \(-\frac{53}{4}\)
As n denotes the numbers of terms, it is a positive integer.
∴ n = \(-\frac{53}{4}\) is not possible.
∴ n = 12
Thus, 12 terms of the AP 9, 17, 25,… must be taken to give a sum of 636.

Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
Here, a = 5; l = 45; S_n = 400; n = ?; d = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ 400 = \(\frac{n}{2}\)(5 + 45)
∴ 800 = n (50)
∴ n = 16
l = a_n = a + (n – 1)d
∴ a₁₆ = a + 15d
∴ 45 = 5 + 15d
∴ 40 = 15d
∴ d = \(\frac{40}{15}\)
∴ d = \(\frac{8}{3}\)
Thus, the number of terms is 16 and the common difference is \(\frac{8}{3}\).

Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
Here, a = 17; l = a_n = 350; d = 9; n = ?; S_n = ?
a_n = a + (n – 1)d
∴ 350 = 17 + (n – 1)9
∴ 333 = 9 (n – 1)
∴ n – 1 = 37 ∴ n = 38
Again, S_n = \(\frac{n}{2}\)(a + l)
∴ S₃₈ = \(\frac{38}{2}\)(17 + 350)
∴ S₃₈ = 19 × 367
∴ S₃₈ = 6973
Thus, there are 38 terms and their sum is 6973.

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Here, a₂₂ = 149; d = 7; S₂₂ = ?
a_n = a + (n – 1) d
∴ a₂₂ = a + 21d
∴ 149 = a + 21 × 7
∴ a = 149 – 147
∴ a = 2
Now, S_n = \(\frac{n}{2}\)(a + l)
∴ S₂₂ = \(\frac{22}{2}\)(2 + 149)
∴ S₂₂ = 11 × 151
∴ S₂₂ = 1661
Thus, the sum of first 22 terms of the given AP is 1661.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Here, a₂ = 14; a₃ = 18; S₅₁ = ?
a_n = a + (n – 1)d
∴ a₂ = a + d = 14 ……..(1)
∴ a₃ = a + 2d = 18 ……..(2)
Solving equations (1) and (2), we get
d = 4 and a = 10.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₅₁ = \(\frac{51}{2}\)[20 + 50 × 4]
∴ S₅₁ = 51 × 110
∴ S₅₁ = 5610
Thus, the sum of first 51 terms of the given AP is 5610.

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Here, S₇ = 49; S₁₇ = 289: S_n = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₇ = \(\frac{7}{2}\)[2a + 6d]
∴ 49 = 7(a + 3d)
∴ a + 3d = 7 ……..(1)
Again S₁₇ = \(\frac{17}{2}\)[2a + 16d]
∴ 289 = 17(a + 8d)
∴ a + 8d = 17 ……..(2)
Solving equations (1) and (2), we get d = 2 and a = 1.
∴ S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S_n = \(\frac{n}{2}\)[2 + (n – 1)2]
∴ S_n = \(\frac{n}{2}\)[2 + 2n – 2]
∴ S_n = \(\frac{n}{2}\)(2n)
∴ S_n = n²
Thus, the sum of first n terms of the given AP is n².

Question 10.
Show that a₁, a₂, ……., a_n, ……. form an AP where a is defined as below:
1. a_n = 3 + 4n
2. a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
1. a_n = 3 + 4n
a₁ = 3 + 4(1) = 7,
a₂ = 3 + 4(2) = 11,
a₃ = 3 + 4(3) = 15,
a₄ = 3 + 4(4) = 19 and so on.
Here, a₄ – a₃ = a₃ – a₂ = a₂ – a₁ = 4.
Hence, a_k+1 – a_k remains the everywhere.
Hence, a₁, a₂, a₃, ….. defined as a_n = 3 + 4n form an AP in which a = 7 and d = 4.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₁₅ = \(\frac{15}{2}\)[14 + 14 × 4]
∴ S₁₅ = 15 × 35
∴ S₁₅ = 525
The sum of first 15 terms of the given AP is 525.

2. a_n = 9 – 5n
a₁ = 9 – 5(1) = 4,
a₂ = 9 – 5(2) = -1,
a₃ = 9 – 5(3) = -6,
a₄ = 9 – 5(4) = -11 and so on.
Here, a₄ – a₃ = a₃ – a₂ = a₂ – a₁ = -5.
Hence, a_k+1 – a_k remains the same everywhere.
Hence, a₁, a₂, a₃, ….. defined as a_n = 9 – 5n form an AP in which a = 4 and d = -5.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₅ = \(\frac{15}{2}\)[8 + 14 (-5)]
∴ S₁₅ = \(\frac{15}{2}\)(-62)
∴ S₁₅ = 15 × (-31)
∴ S₁₅ = -465
The sum of first 15 terms of the given AP is -465.

Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
For the given AP
S_n = 4n – n²
∴ S₁ = 4(1) – (1)² = 4 – 1 = 3,
S₂ = 4 (2) – (2)² = 8 – 4 = 4,
S₃ = 4(3) – (3)² = 12 – 9 = 3,
S₉ = 4 (9) – (9)² = 36 – 81 = -45,
S₁₀ = 4 (10) – (10)² = 40 – 100 = -60
Now, the first term = a – a₁ = S₁ = 3
The sum of first two terms S₂ = 4
The second term a₂ = S₂ – S₁ = 4 – 3 = 1
The third term a₃ = S₃ – S₂ = 3 – 4 = -1
The tenth term a₁₀ = S₁₀ – S₉
= -60 – (-45) = -15
Now, S_n = 4n – n²
∴ S_n-1 = 4(n – 1) – (n – 1)²
= 4n – 4 – n² + 2n – 1
= -n² + 6n – 5
Now nth term a_n = S_n – S_n-1
∴ a_n = (4n – n²) – (-n² + 6n – 5)
∴ a_n = 4n – n² + n² – 6n + 5
∴ a_n = -2n + 5

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The first 40 positive integers divisible by 6 form the AP 6, 12, 18, ……, 240.
Here, a = 6; d = 12 – 6 = 6; n = 40 and l = 240.
S_n = \(\frac{n}{2}\)(a + l)
∴ S₄₀ = \(\frac{40}{2}\)(6 + 240)
∴ S₄₀ = 20 × 246
∴ S₄₀ = 4920
Thus, the required sum is 4920.

Alternate method:
Required sum
= 6 + 12 + 18 + … + 240
= 6(1 + 2 + 3 + … + 40)
= 6 × \(\frac{40 \times 41}{2}\) (1 + 2 + 3 + …. + n = \(\frac{n(n+1)}{2}\))
= 4920

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 form the AP 8, 16, 24, ….., 120.
Here, a = 8, d = 16 – 8 = 8, n = 15 and l = 120.
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₅ = \(\frac{15}{2}\)(8 + 120)
∴ S₁₅ = 15 × 64
∴ S₁₅ = 960
Thus, the required sum is 960.

Alternate method:
Required sum
= 8 + 16 + 24 + … + 120
= 8(1 + 2 + 3 + … + 15)
= 8 × \(\frac{15 \times 16}{2}\) (1 + 2 + 3 + … + n = \(\frac{n(n+1)}{2}\))
= 960.

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 form the AP 1, 3, 5, ….., 49.
Here, a = 1, d = 3 – 1 = -2, l = 49.
Let the last term be the nth term.
a_n = a + (n – 1) d
49 = 1 + (n – 1)²
2(n – 1) = 48
n – 1 = 24
n = 25
Now, S_n = \(\frac{n}{2}\)(a + l)
∴ S₂₅ = \(\frac{25}{2}\)(1 + 49)
∴ S₂₅ = 25 × 25
∴ S₂₅ = 625
Thus, the required sum is 625.

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc.. the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
The sums (in rupees) of penalty for delay of completion form the AP 200, 250, 300, …..
Here, a = 200; d = 250 – 200 = 50; n = 30 as the contractor has delayed the work by 30 days. The total penalty amount (in rupees) to be paid by the contractor will be given by S₃₀.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₃₀ = \(\frac{30}{2}\)[400 + (30 – 1)50]
∴ S₃₀ = 15 × 1850
∴ S₃₀ = 27750
Thus, the contractor has to pay a penalty of ₹ 27,750.

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If cach prize is 20 less than its preceding prize, find the value of each of the prizes.
Solution:
₹ 700 is to be distributed as seven prizes such that each prize is ₹ 20 less than its preceding prize. Let the highest prize, i.e., the first prize be ₹ a. Then, the second prize will be of ₹ a – 20, the third prize will be of ₹ a – 40 and so on up to seven prizes. Hence, the amount (in rupees) of these prizes form a finite AP with seven terms as a, a – 20, a – 40, a – 60, a – 80, a – 100 and a – 120.
Here, the first term = a; d = (a – 20) – a = -20;
n = 7 and the sum of all the terms = S₇ = 700.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 700 = \(\frac{7}{2}\)[2a + (7 – 1) (-20)]
∴ 200 = 2a + 6(-20)
∴ 200 = 2a – 120
∴ 2a = 320
∴ a = 160
Then, a – 20 = 140; a – 40 = 120; a – 60 = 100; a – 80 = 80; a – 100 = 60 and a – 120 = 40.
Thus, the values (in rupees) of those seven prizes are 160, 140, 120, 100, 80, 60 and 40.

Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class. will plant, will be the same as the class, in which they are studying, e.g.. a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
The number of trees that the three sections of Class I will plant = 1 + 1 + 1 = 3.
The number of trees that the three sections of Class II will plant = 2 + 2 + 2 = 6.
This system will continue till Class XII.
The number of trees that the three sections of Class XII will plant = 12 + 12 + 12 = 36.
Thus, the number of trees that will be planted will form a finite AP with 12 terms as 3, 6, 9, ….., 36.
Here, a = 3, d = 6 – 3 = 3, n = 12 and S₁₂ will give the total number of trees that will be planted.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₂ = \(\frac{12}{2}\)[6 + (12 – 1)³]
∴ S₁₂ = 6 × 39
∴ S₁₂ = 234
Thus, 234 trees will be planted by the students.

Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, as shown in the given figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = \(\frac{22}{7}\))

Hint: Length of successive semicircles is l₁, l₂, l₃, l₄, …with centres at A, B, A, B, ….., respectively.]
Solution:
We know that the length of a semicircle = πr, where r is the radius.
Length of 1st semicircle with centre A and radius 0.5 cm = l₁ = π × 0.5 cm.
Length of 2nd semicircle with centre B and radius 1 cm = l₂ = π × 1 cm.
Length of 3rd semicircle with centre A and radius 1.5 cm = l₃ = π × 1.5 cm.
This system continues till 13 semicircles are drawn.
Then, the 13th semicircle will be drawn with centre A and radius 6.5 cm. Length of 13th semicircle with centre A and radius 6.5 cm = l₁₃ = π × 6.5 cm.
Now, the total length of the spiral
= l₁ + l₂ + l₃ + … + l₁₃
= (π × 0.5) + (π × 1) + (π × 1.5) + … + (π × 6.5)
= π(0.5 + 1 + 1.5 + … + 6.5)
The sum inside the brackets is the sum of all the 13 terms of the finite AP 0.5, 1, 1.5, ….., 6.5.
For this AP a = 0.5; d = 1 – 0.5 = 0.5 and n = 13.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S_n = [1 + (13 – 1) (0.5)]
∴ S_n = \(\frac{13}{2}\) × 7
Hence, the total length of the spiral
= π\(\left(\frac{13}{2} \times 7\right)\)
= \(\frac{22}{7} \times \frac{13}{2} \times 7\)
= 143 cm
Thus, the total length of the spiral made up of thirteen consecutive semicircles is 143 cm.

Question 19.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see the given figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:
The number of logs stacked in the first row from the bottom = 20.
The number of logs stacked in the second row from the bottom = 19.
The number of logs stacked in the third row from the bottom = 18.
This system continues till all the 200 logs are stacked.
Thus, the number of logs stacked in the rows form the finite AP 20, 19, 18,….. up ton-terms and the sum of those n terms is 200. Here, a = 20 and d = 19 – 20 = -1.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 200 = \(\frac{n}{2}\)[40 + (n – 1) (-1)]
∴ 400 = n(40 – n + 1)
∴ 400 = n(41 – n)
∴ 400 = 41n – n²
∴ n² – 41n + 400 = 0
∴ n² – 16n – 25n + 400 = 0
∴ n (n – 16) – 25 (n – 16) = 0
∴ (n – 16) (n – 25) = 0
n – 16 = 0 or n – 25 = 0
n = 16 or n = 25
Here, both the answers are admissible. Hence, we verify by the value of 16th term and 25th term.
a_n = a + (n – 1) d
∴ a₁₆ = 20 + 15(-1) = 5
∴ a₂₅ = 20 + 24(-1) = -4
Thus, for the 25th row, the number of logs in the row becomes negative. This is inadmissible.
Hence, n ≠ 25.
∴ n = 16 and a₁₆ = 5.
Thus, the 200 logs are placed in 16 rows and in the top row there are 5 logs.

Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see the given figure).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)].
Solution:
The distance (in metres) to be covered to pick up first potato = 2 × 5 = 10.
The distance (in metres) to be covered to pick up second potato = 2 × (5 + 3) = 16.
The distance (in metres) to be covered to pick up third potato = 2 × (5 + 3 + 3) = 22.
Thus, the distances to be covered to pick up 10 potatoes form the finite AP 10, 16, 22, …. up to 10 terms.
Here, a = 10, d = 16 – 10 = 6, n = 10 and Sn will give the total distance (in metres) that the competitor has to run.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀ = \(\frac{10}{2}\)[20 + (10 – 1)6]
∴ S₁₀ = 5 × 74
∴ S₁₀ = 370
Thus, the total distance that the competitor has to run is 370 metres.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1

Question 1.
In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A, (ii) sin C, cos C.
Solution :

(i) sin A, cos A
AC² = AB² + BC²
= 24² + 7² = 576 + 49
= 625
∴ AC = 25

Question 2.
In the figure, find tan P – cot R.
Solution :

QR² = PR² – PQ² = 13² – 12²
= 169 – 144 = 25.
∴ QR = 5.
tan P = \(\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{5}{12}\)
cot R = \(\frac{\mathrm{QR}}{\mathrm{QP}}=\frac{5}{12}\)
∴ tan P – cot R = \(\frac{5}{12}=\frac{5}{12}\) = 0

Question 3.
If sin A = \(\frac{3}{4}\), calculate cos A and tan A.
Solution:

Given, sin A = \(\frac{3}{4}\) ⇒ \(\frac{P}{H}\) = \(\frac{3}{4}\)
Let P = 3k and H = 4k.
In right angled ΔABC,
P² + B² = H² (By Pythagoras theorem)
⇒ (3k)² + B² = (4k)²
⇒ 9k² + B² = 16k²
⇒ B² = 16k² – 9k² = 7k²
∴ B = +k\(\sqrt{7}\) (∵ Base cannot be -ve)

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:

Given, 15 cot A = 8 ⇒ cot A = \(\frac{8}{15}\)
\(\frac{\mathrm{B}}{\mathrm{P}}=\frac{8}{15}\)
B = 8k and P = 15 k
In right angled ΔABC, H² = B² + P² (By Pythagoras theorem)
= (8k)² + (15k)²
= 64k² + 225k²
= 289k²
∴ H = 17k (∵ Side cannot be -ve)
Now, sin A = \(\frac{\mathrm{P}}{\mathrm{H}}=\frac{15 \mathrm{k}}{17 \mathrm{k}}=\frac{15}{17}\)
sec A = \(\frac{\mathrm{H}}{\mathrm{B}}=\frac{17 \mathrm{k}}{8 \mathrm{k}}=\frac{17}{8}\)

Question 5.
Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios.
Solution :

sec θ = \(\frac{1}{cos θ}\)
sec θ . cos θ = 1
\(\frac{13}{12} \times \frac{12}{13}\) = 1
AB² = AC² – BC²
= 13² – 12²
= 169 – 144
= 25
AB = \(\sqrt{25}\) = 5

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution :

cos A = \(\frac{AC}{AB}\)
cos B = \(\frac{BC}{AB}\)
cos A = cos B
\(\frac{AC}{AB}=\frac{BC}{AB}\)
∴ AC = BC
∴ \(\hat{A}\) = \(\hat{B}\) (∵ Angles opposite to equal sides are also equal)

Question 7.
If cot θ = \(\frac{7}{8}\), evaluate: (i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\) (ii) cot² θ
Solution :

AC² = AB² + BC²
= 7² + 8²
= 49 + 64 = 113
AC = \(\sqrt{113}\)

(ii) cot² θ = (cot θ)²
= (\(\frac{7}{8}\))² = \(\frac{49}{64}\)
Given cot θ = \(\frac{7}{8}\)

Question 8.
If 3 cot A = 4, check whether \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\) = cos² A – sin² A or not.
Solution :

3 cot A = 4
cot A = \(\frac{4}{3}\)
AC² = AB² + BC²
= 4² + 3²
= 16 + 9
= 25
AC = \(\sqrt{25}\) = 5

Question 9.
In triangle ABC, right-angled at B, if tan A = find the value of :
(i) sin A cos C + cos A sin C,
(ii) cos A cos C – sin A sin C.
Solution :
(i) sin A cos C + cos A sin C

(ii) cos A . cos C – sin A . sin C
\(\frac{AB}{AC} \cdot \frac{BC}{AC}\) – \(\frac{BC}{AC} \cdot \frac{AB}{AC}\)

Question 10.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
PR + QR = 25
QR = 25 – PR
PQ² + QR² = PR²
(5)² + (25 – PR)² = PR²
25 + (625 – 50PR + PR²) = PR²
650 – 50PR = PR² – PR² = 0
– 50PR = – 650
PR = \(\frac{-650}{-50}\) = 13.
PR = 13, QR = 25 – PR = 25 – 13 = 12, PQ = 5.

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = \(\frac{12}{5}\) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = \(\frac{4}{3}\) for some angle θ.
Solution :
(i) False, because the value of tan A increases from 0 to ∞. Also, tan 45° = 1.
(ii) True, because the value of sec A increases from 1 to ∞.
(iii) False, cos A is the abbreviation used for the cosine of angle A.
(iv) False, because cot A is one symbol. We cannot separate cot and A.
(v) False, because the value of sin θ always lies between 0 and 1. Here, sin θ = \(\frac{4}{3}\) which is greater than 1. So, it is not possible.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and a, the nth term of the AP:

Solution:
1. Here, a = 7, d = 3, n = 8 and a_n is to be found.
We have a_n = a + (n – 1)d
a₈ = 7 + (8 – 1) 3 = 7 + 21 = 28

2. Here, a = -18, n = 10, a_n = a₁₀ = 0 and d is to be found.
a_n = a + (n – 1)d
∴ 0 = – 18 + (10 – 1)d
∴ 18 = 9d ∴ d = 2

3. Here, d = -3, n = 18, a_n = a₁₈ = -5 and a is to be found.
an = a + (n – 1)d
∴ -5 = a + (18 – 1)(-3)
∴ -5 = a – 51
∴ a = 51 – 5 ∴ a = 46

4. Here, a = -18.9, d = 2.5, a_n = 3.6 and n is to be found.
a_n = a + (n – 1)d
∴ 3.6 = – 18.9 + (n – 1)(2.5)
∴ 22.5 = 2.5 (n – 1)
∴ (n – 1) = \(\frac{22.5}{2.5}\)
∴ n – 1 = 9 ∴ n = 10

5. Here, a = 3.5, d = 0, n = 105 and a_n is to be found.
a_n = a + (n – 1) d
∴ a₁₀₅ = 3.5 + (105 – 1) (0)
∴ a₁₀₅ = 3.5

Question 2.
Choose the correct choice in the following and justify:
1. 30th term of the AP: 10, 7, 4, ……, is
(A) 97
(B) 77
(C) -77
(D) -87
2. 11th term of the AP: -3, –\(\frac{1}{2}\), 2, ….. is
(A) 28
(B) 22
(C) -38
(D) -48\(\frac{1}{2}\)
Solution:
1. For the given AP 10, 7, 4,… a = 10,
d = 7 – 10 = -3 and n = 30
a_n = a + (n – 1)d
∴ a₃₀ = 10 + (30 – 1) (-3)
∴ a₃₀ = 10 – 87
∴ a₃₀ = -77
Thus, the correct choice is (C) -77.

2. For the given AP -3, –\(\frac{1}{2}\), 2, ….., a = -3.
d = –\(\frac{1}{2}\) – (-3) = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) and n = 11.
a_n = a + (n – 1)d
∴ a₁₁ = -3 + (11 – 1)(\(\frac{5}{2}\))
∴ a₁₁ = – 3 + 25
∴ a₁₁ = 22
Thus, the correct choice is (B) 22.

Question 3.
In the following APs, find the missing terms in the boxes:

Solution:
For the given AP, first term = a = 2 and third term = a + 2d = 26.
a = 2 and a + 2d = 26 gives d = 12.
Then, second term = a + d = 2 + 12 = 14
Thus, the missing term in the box is

2. For the given AP,
second term = a + d = 13 …….(1)
fourth term = a + 3d = 3 …….(2)
Solving equations (1) and (2) we get d = -5 and a = 18.
Now, first term = a = 18 and third term = a + 2d
= 18 + 2(-5) = 8.
Thus, the missing terms in the boxes are

Alternative Method:
Let the terms of the given AP be a₁, a₂, a₃, a₄.
Here, a₂ = 13 and a₄ = 3.
Now, a₄ – a₃ = a₃ – a₂ = d
∴ 3 – a₃ = a₃ – 13
∴ 2a₃ = 16
∴ a₃ = 8
Again, a₂ – a₁ = a₃ – a₂
∴ 13 – a₁ = 8 – 13
∴ 13 – a₁ = -5
∴ a₁ = 18
Thus, the missing terms in the boxes are

3. For the given AP
first term a = 5 ……….(1)
fourth term = a + 3d = 9\(\frac{1}{2}\) ……….(2)
From equations (1) and (2), we get a = 5 and d = 1\(\frac{1}{2}\).
Now, second term = a + d = 5 + 1\(\frac{1}{2}\) = 6\(\frac{1}{2}\)
and third term = a + 2d = 5 + 2 (1\(\frac{1}{2}\)) = 8
Thus, the missing terms in the boxes are

4. For the given AP
first term a = -4 ……….(1)
sixth term = a + 5d = 6 ……….(2)
From equations (1) and (2), we get
a = -4 and d = 2
Now, second term = a + d(-4) + 2 = -2,
third term = a + 2d = (-4) + 2 (2) = 0.
fourth term = a + 3d = (-4) + 3(2) = 2 and
fifth term = a + 4d = (-4) + 4(2) = 4.
Thus, the missing terms in the boxes are

5. For the given AP,
second term = a + d = 38 ……….(1)
sixth term = a + 5d = -22 ……….(2)
Solving equations (1) and (2), we get
d = -15 and a = 53.
Now, first term = a = 53,
third term = a + 2d = 53 + 2(-15) = 23,
fourth term = a + 3d = 53 + 3(-15) = 8 and
fifth term = a + 4d = 53 + 4(-15) = -7.
Thus, the missing terms in the boxes are

Question 4.
Which term of the AP: 3, 8, 13, 18, … is 78 ?
Solution:
Suppose nth term of the AP 3, 8, 13, 18, … is 78.
Here, a = 3, d = 8 – 3 = 5, a_n = 78 and n is to be found.
a_n = a + (n – 1)d
∴ 78 = 3 + (n – 1)5
∴ 75 = 5 (n-1)
∴ 15 = n – 1 ∴ n = 16
Thus, the 16th term of the AP 3, 8, 13, 18, …….., is 78.

Question 5.
Find the number of terms in each of the following APs:
1. 7, 13, 19, …….., 205
2. 18, 15\(\frac{1}{2}\), 13, ……., -47
Solution:
1. For the given finite AP 7, 13, 19, ….. 205, a = 7, d = 13 – 76 and last term l = 205.
Let us consider that the last term is the nth term.
a_n = a + (n – 1)d
∴ 205 = 7 + (n – 1)6
∴ 198 = 6 (n – 1)
∴ n – 1 = 33
∴ n = 34
Thus, there are 34 terms in the given finite AP.

2. For the given finite AP 18, 15\(\frac{1}{2}\), 13….. -47, a = 18, d = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\) and last term l = -47.
Let us consider that the last term is the nth term.
a_n = a + (n – 1) d
∴ -47 = 18 + (n-1) (-\(\frac{5}{2}\))
∴ -65 = –\(\frac{5}{2}\)(n – 1)
∴ n – 1 = 26
∴ n = 27
Thus, there are 27 terms in the given finite AP.

Question 6.
Check whether -150 is a term of the AP: 11, 8, 5, 2…
Solution:
If possible, let -150 be the nth term of the AP 11, 8, 5, 2,…
Here, a = 11; d = 8 – 11 = -3 and a_n = -150.
a_n = a + (n – 1)d
∴ -150 = 11 + (n – 1)(-3)
∴ -161 = -3 (n – 1)
∴ n – 1 = \(\frac{161}{3}\)
∴ n = \(\frac{164}{3}\)
∴ But n must be positive integer for an AP.
Hence, -150 cannot be a term of the AP 11, 8, 5, 2…..

Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
For any AP, a_n = a + (n – 1)d.
a₁₁ = a + 10 d
∴ a + 10 d = 38 ………(1)
∴ a₁₆ = a + 15d
∴ a + 15d = 73 ………(2)
Solving equations (1) and (2), we get
d = 7 and a = -32.
Now, 31st term = a₃₁ = a + 30d
= -32 + 30(7)
= -32 + 210
= 178
Thus, the 31st term of the given AP is 178.
Note: d = \(\frac{a_{16}-a_{11}}{16-11}=\frac{73-38}{5}=\frac{35}{5}=7\) can also be used.

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
The given finite AP has 50 terms, hence its last term is a₅₀.
So, a₃ = 12 and a₅₀ = 106.
Now, a_n = a + (n – 1)d.
That gives, a₃ = a + 2d = 12 ………(1)
and a₅₀ = a + 49d = 106 ………(2)
Solving equations (1) and (2), we get
d = 2 and a = 8.
Now, 29th term = a₂₉ = a + 28d
∴ a₂₉ = 8 + 28 (2)
∴ a₂₉ = 64
Thus, the 29th term of the given AP is 64.

Question 9.
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
For the given AP, a₃ = 4 and a9 = -8
We know, a_n = a + (n – 1)d.
∴ a₃ = a + 2d = 4 ………(1)
and a₉ = a + 8d = -8 ………(2)
Solving equations (1) and (2), we get
d = -2 and a = 8.
Now, let nth term of the AP be 0.
a_n = a + (n – 1)d
∴ 0 = 8 + (n – 1) (-2)
∴ 2(n – 1) = 8
∴ n – 1 = 4
∴ n = 5
Thus, the 5th term of the given AP is zero.

Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
For the given AP,
a₁₇ + 16d = a + 9d + 7a [∵ a_n = a + (n – 1) d]
∴ 7d = 7
∴ d = 1.
Thus, the common difference of the given AP is 1.

Question 11.
Which term of the AP:3, 15, 27, 39, …….. will be 132 more than its 54th term?
Solution:
For the given AP 3, 15, 27, 39, …….., a = 3
and d = 15 – 3 = 12.
Suppose nth term of the AP is 132 more than its 54th term.
∴ a_n = a₅₄ + 132
∴ a + (n – 1)d = a + 53d + 132
∴ 3 + (n – 1) (12) = 3 + 53(12) + 132
∴ 12(n – 1) = 12 (53 + 11)
∴ 12(n – 1) = 12 × 64
∴ n – 1 = 64
∴ n = 65
Thus, the 65th term of the given AP is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let the first term of two given APs be a₁ and a₂ respectively and let the same common difference be d.
Also, let a₁ > a₂
Then, 100th term of the first AP = a₁ + 99d
(a_n = a + (n – 1)d)
100th term of the second AP = a₂ + 99d.
The difference between their 100th terms is 100.
∴ (a₁ + 99d) – (a₂ + 99d) = 100 (a₁ > a₂)
∴ a₁ – a₂ = 100 ……(1)
Now, 1000th term of the first AP = a₁ + 999d
1000th term of the second AP = a₂ + 999d.
Then, the difference between their 1000th terms
= (a₁ + 999d) – (a₂ + 999d)
= a₁ – a₂
= 100 (by (1))
Thus, the difference between the 1000th terms of the two APs is 100.

Question 13.
How many three digit numbers are divisible by 7?
Solution:
The list of three digit numbers divisible by 7 is as below:
105, 112, 119, …….., 987, 994.
These numbers form a finite AP with a = 105, d = 112 – 105 = 7 and last term l = 994.
Suppose the last term of the AP is its nth term.
∴ l = a_n
∴ 994 = a + (n – 1)d
∴ 994 = 105 + (n – 1)7
∴ 7(n – 1) = 889
∴ n – 1 = 127
∴ n = 128
Hence, there are 128 terms in the AP.
Hence, 128 three digit numbers are divisible by 7.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The multiples of 4 lying between 10 and 250 give rise to following finite AP:
12, 16, 20, ….., 244, 248.
Here, a = 12, d = 16 – 12 = 4 and last term l = 248.
If the last term is the nth term of the AP then l = a_n.
∴ l = a + (n – 1) d
∴ 248 = 12 + (n – 1)4
∴ 236 = 4 (n – 1)
∴ n – 1 = 59
∴ n = 60
Thus, there are 60 terms in the AP.
Thus, 60 multiples of 4 lie between 10 and 250.

Question 15.
For what value of n, are the nth terms of two APs 63, 65, 67, …… and 3, 10, 17,… equal?
Solution:
For the first AP 63, 65, 67, ….., a = 63, d = 65 – 63 = 2.
Then, nth term of the first AP a is given by a_n = a + (n – 1)d = 63 + (n – 1)(2).
For the second AP 3, 10, 17, ……, A = 3, D = 10 – 3 = 7.
Then, nth term of the second AP An is given by
A_n = A + (n – 1) D = 3 + (n – 1) (7).
Now, a_n = A_n
∴ 63 + (n – 1) (2) = 3 + (n – 1)(7)
∴ 63 – 3 = (n – 1)(7 – 2)
∴ 60 = 5(n – 1)
∴ n – 1 = 12
∴ n = 13
Thus, for n = 13, the nth term of two given APs are equal.

Question 16.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
For the given AP a₃ = 16 and a₇ = a₅ + 12.
For any AP, a_n = a + (n – 1) d.
∴ a + 2d = 16 and a + 6d = a + 4d + 12
a + 6d = a + 4d + 12 gives 2d = 12,
i.e., d = 6.
Substituting d = 6 in a + 2d = 16, we get a = 4.
Then, the required AP is 4, 4 + 6, 4 + 2 (6), 4 + 3(6), …..
Hence, the required AP is 4, 10, 16, 22, ……

Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, ….., 253.
Solution:
For the given finite AP 3, 8, 13, …., 253,
a = 3, d = 8 – 3 = 5 and last term l = 253.
Let the last term be its nth term.
∴ l = a_n
∴ l = a + (n – 1)d
∴ 253 = 3 + (n – 1)(5)
∴ 250 = 5 (n – 1)
∴ n – 1 = 50
∴ n = 51
Thus, there are in all 51 terms in the AP.
Now, the 20th term from the last term is (51 – 20 + 1)th term = 32nd term from the beginning.
a₃₂ = a + 31d
∴ a₃₂ = 3 + 31(5)
∴ a₃₂ = 158
Thus, the 20th term from the last term is 158.

Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
For any AP, a_n = a + (n – 1) d.
a₄ = a + 3d, a₈ = a + 7d, a₆ = a + 5d and a₁₀ = a + 9d.
Now, a₄ + a₈ = 24 (Given)
∴ (a + 3d) + (a + 7d) = 24
∴ 2a + 10d = 24
∴ a + 5d = 12 ……..(1)
Again, a₆ + a₁₀ = 44 (Given)
∴ (a + 5d) + (a + 9d) = 44
∴ 2a + 14d = 44
∴ a + 7d = 22 …….(2)
Solving equations (1) and (2), we get d = 5 and a = -13.
Then, a₂ = a + d = 13 + (5) = -8 and
a₃ = a + 2d = -13 + 2(5) = -3.
Thus, the first three terms of the AP are -13, 8, 3.

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5,000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7,000?
Solution:
Subba Rao’s income in first year = ₹ 5,000
His income in second year = ₹ 5,000 + ₹ 200
= ₹ 5200
His income in third year = ₹ 5200 + ₹ 200
= ₹ 5400
and so on.
These numbers of his income (in rupees) form the AP 5000, 5200, 5400, …….
Here, a = 5000; d = 5200 – 5000 = 200;
a_n = 7000 and n is to be found.
a_n = a + (n – 1)d
∴ 7000 = 5000 + (n – 1)(200)
∴ 2000 = 200(n – 1)
∴ n – 1 = 10
∴ n = 11
Thus, Subba Rao’s income will reach ₹ 7000 in 11th year, i.e., in the year 2005.

Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Solution:
Ramkali’s savings in first week = ₹ 5
her savings in second week = ₹ 5 + ₹ 1.75
= ₹ 6.75,
her savings in third week = ₹ 6.75 + ₹ 1.75
= ₹ 8.50.
and so on.
Thus, the weekly savings (in rupees) of Ramkalt form the AP 5, 6.75, 8.50, …..
Here, a = 5; d = 6.75 – 5 = 1.75; a_n = 20.75 and n is to be found.
a_n = a + (n – 1)d
∴ 20.75 = 5 + (n – 1)(1.75)
∴ 1.75(n – 1) = 15.75
∴ n – 1 = 9
∴ n = 10
Thus, if Ramkali’s weekly savings is ₹ 20.75 in nth week, then n = 10.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.4

Question 1.
Which term of the AP 121, 117, 113, …. is its first negative term?
[Hint: Find n for a_n < 0]
Solution :
For the given AP 121, 117, 113, … a = 121 and d= 117 – 121 = – 4.
Let n^th term of the AP be its first negative term.
∴ a_n < 0
∴ a + (n – 1) d < 0
∴ 121 + (n – 1)(- 4) < 0
∴ 121 < 4 (n – 1) ∴ n > \(\frac{125}{4}\)
∴ n > 31\(\frac{1}{4}\)
Now, n being the number of a term is a positive integer and the smallest positive integer satisfying n > 31\(\frac{1}{4}\) is 32.
Hence, the 32nd term of the given AP is its first negative term.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Solution :
For the given AP, let the first term be a and the common difference be d.
a_n = a + (n – 1) d
∴ a₃ = a + 2d and a₇ = a + 6d
According to given information,
a₃ + a₇ = 6
∴ (a + 2d) + (a + 6d) = 6
∴ 2a + 8d = 6
∴ a + 4d = 3
∴ a = 3 – 4d …………….(1)
Again, the product of a₃ and a₇ is 8.
∴ (a + 2d) (a + 6d) = 8
∴ (3 – 4d + 2d) (3 – 4d + 6d) = 8 [by (1)]
∴ (3 – 2d) (3 + 2d) = 8
∴ 9 – 4d² = 8
∴ 1 = 4d²
∴ d² = \(\frac{1}{4}\)
∴ d = \(\frac{1}{2}\) or d = – \(\frac{1}{2}\)

Thus, the required sum of first sixteen terms is 76 or 20.

Question 3.
A ladder has rungs 25 cm apart. (see the given figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2\(\frac{1}{2}\)m apart, what is the length of the wood required for the rungs?
Solution :

The distance between the top rung and the bottom rung = 2\(\frac{1}{2}\)m = 250 cm.
The distance between two successive rungs = 25 cm.
∴ Total number of rungs = \(\frac{250}{25}\) + 1 = 11
Including the top rung as well as the bottom rung
The length of the first (bottom) rung = 45 cm.
The length of the 11th rung at top = 25 cm.
The length of rung decreases uniformly.
Hence, the lengths (in cm) of rungs form an AP in which the first term = 45 and 11th term = 25.
a_n = a+ (n – 1)d
∴ a₁₁ = a + 10 d
∴ 25 = 45 + 10 d
∴ – 20 = 10 d
∴ d = – 2
Thus, the length of rung uniformly decreases by 2 cm as we move from bottom to top.
Thus, the lengths (in cm) of rungs form a finite AP 45, 43, 41, ……….with 11 terms.
Then, sum of all the eleven terms will give the total length of the wood required for the rungs.
S_n = \(\frac{n}{2}\)(a + 1)
∴ S₁₁ = \(\frac{11}{2}\)(45 + 25)
∴ S₁₁ = \(\frac{11}{2}\) × 70
∴ S₁₁ = 385
Thus, the total length of the wood required for the rungs is 385 cm.

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint: S_x-1 = S₄₉ – S_x]
Solution :
We know the sum of first n positive integers n(n+1) is given by S_n = \(\frac{n(n+1)}{2}\)
According to data, 1 + 2 + 3 + ……… + (x – 1) = (x + 1) + (x + 2) + ……….. + 49
∴ \(\frac{(x-1) \cdot x}{2}\) = (1 + 2 + 3 + … + 49) – (1 + 2 + 3 + … + x)
[Adding and subtracting (1 + 2 + 3 + … + x)}
∴ \(\frac{(x-1)(x)}{2}=\frac{49 \times 50}{2}-\frac{x(x+1)}{2}\)
∴ x(x – 1) + x(x + 1) = 49 × 50
∴ x² – x + x² + x = 49 × 50
∴ 2x² = 49 × 50
∴ x² = \(\frac{49 \times 50}{2}\)
∴ x² = 49 × 25
∴ x = 7 × 5
∴ x = 35
Thus, the value of x is 35.

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see the given figure). Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the first step = \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m³]
Solution :

Volume of concrete required to build the first step = 50 × \(\frac{1}{2}\) × \(\frac{1}{4}\)m³ = \(\frac{25}{4}\)m³
Volume of concrete required to build the second step = 50 × \(\frac{1}{2}\) × (\(\frac{1}{4}\) + \(\frac{1}{4}\))m³ = \(\frac{25}{2}\)m³
Volume of concrete required to build the third step = 50 × \(\frac{1}{2}\) × (\(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\))m³ = = \(\frac{75}{4}\)m³ and so on up to 15 steps.
Thus, the volumes (in m³) of concrete required to build those 15 steps form the finite AP \(\frac{25}{4}\), \(\frac{25}{2}\), \(\frac{75}{4}\) …with 15 terms.
The sum of all the fifteen terms will give the quantity of total concrete required.
Here, a = \(\frac{25}{4}\), d = \(\frac{25}{2}-\frac{25}{4}=\frac{25}{4}\) and n = 15.
S_n = \(\frac{n}{2}\)[(2a + (n – 1) d]
∴ S₁₅ = \(\frac{15}{2}\)[\(\frac{25}{2}\) +(15 – 1)\(\frac{25}{4}\)]
∴ S₁₅ = \(\frac{15}{2}\) [latex]\frac{25}{2}+\frac{175}{2}[/latex]
∴ S₁₅ = \(\frac{15}{2} \times \frac{200}{2}\)
∴ S₁₅ = 15 × 50
∴ S₁₅ = 750
Thus, 750 m³ of concrete is required to build the terrace.