# JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2

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 an is to be found.
We have an = a + (n – 1)d
a8 = 7 + (8 – 1) 3 = 7 + 21 = 28

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

3. Here, d = -3, n = 18, an = 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, an = 3.6 and n is to be found.
an = 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 an is to be found.
an = 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
an = 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.
an = 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, an = 78 and n is to be found.
an = 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.
an = 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.
an = 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 an = -150.
an = 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, an = 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, an = 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, an = 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.
an = 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 [∵ an = 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.
∴ an = 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
(an = 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 = an
∴ 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 = an.
∴ 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 an = 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
An = A + (n – 1) D = 3 + (n – 1) (7).
Now, an = An
∴ 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, an = 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 = an
∴ 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, an = 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;
an = 7000 and n is to be found.
an = 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; an = 20.75 and n is to be found.
an = 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.