JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
1. The taxi fare after each km, when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
2. The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
3. The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
4. The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at 8% per annum.
Solution:
1. Here, the fare for 1 km = ₹ 15
the fare for 2 km = ₹ 15 + ₹ 8 = ₹ 23,
the fare for 3 km = ₹ 15 + 2 (₹8) = ₹ 31,
the fare for 4 km = ₹ 15 + 3 (₹8) = ₹ 39, and so on.
The list of numbers formed is 15, 23, 31, 39, …………..
Here, a₂ – a₁ = 23 – 15 = 8.
a₃ – a₂ = 31 – 23 = 8.
a₄ – a₃ = 39 – 31 = 8, and so on.
Thus, a_k+1 – a_k is the same every time.
Hence, the list of numbers forms an AP with a = 15 and d = 8.

2. Let the volume of air present in the cylinder at the beginning be V units. Then, volume of air remaining in the cylinder after first attempt = \(\frac{3}{4}\)V units. Also, volume of air remaining in the cylinder after second attempt = \(\left(\frac{3}{4}\right)^2\)V units. Here, the list of numbers formed is V, \(\frac{3}{4}\)V, \(\left(\frac{3}{4}\right)^2\)V, ………
Now, a₂ – a₁ = \(\frac{3}{4}\)V – V = –\(\frac{1}{4}\)V
a₃ – a₂ = \(\left(\frac{3}{4}\right)^2\)V – \(\frac{3}{4}\)V
= \(V\left(\frac{9}{16}-\frac{3}{4}\right)\)
= \(– \frac{3}{16}\)V
Here, a₂ – a₁ ≠ a₃ – a₂.
Hence, the list of numbers does not form an AP.

3. Cost of digging first metre = ₹ 150
Cost of digging the second metre = ₹ 150 + ₹ 50
= ₹ 200
Cost of digging the third metre = ₹ 200 + ₹ 50
= ₹ 250
Cost of digging the fourth metre = ₹ 250 + ₹ 50
= ₹ 300
The list of numbers formed is 150, 200, 250, 300,…
Here, a₂ – a₁ = 200 – 150 = 50,
a₃ – a₂ = 250 – 200 = 50,
a₄ – a₃ = 300 – 250 = 50, and so on.
Thus, a_k+1 – a_k is the same every time. Hence, the list of numbers forms an AP with a = 150 and d = 50.

4. The formula of compound interest is known to us.
A = \(P\left(1+\frac{R}{100}\right)^T\)
Here, P = ₹ 10,000; R = 8% and T = 1, 2, 3, 4, ……….
Amount at the end of 1st year = ₹ 10000 (1.08).
Amount at the end of 2nd year = ₹ 10000 (1.08)².
Amount at the end of 3rd year = ₹ 10000 (1.08)³.
The list of numbers formed is 10000 (1.08), 10000 (1.08)2, 10000 (1.08), ………..
a₂ – a₁ = 10000 (1.08)² – 10000 (1.08)³
= 10000 (1.08) (1.08 – 1)
= 10000 (1.08) (0.08)
a₃ – a₂ = 10000 (1.08)³ – 10000 (1.08)²
= 10000 (1.08)² (1.08 – 1)
= 10000 (1.08)² (0.08)
Thus, a₂ – a₁ ≠ a₃ – a₂
Hence, the list of numbers does not form an AP.

Question 2.
Write first four terms of the AP when the first term a and the common difference d are given as follows:
1. a = 10, d = 10
2. a = -2, d = 0
3. a = 4, d = -3
4. a = -1, d = \(\frac{1}{2}\)
5. a = -1.25, d = -0.25
Solution:
1. a = 10, d = 10
First term a = 10
Second term = First term + d
= 10 + 10 = 20
Third term = Second term + d
= 20 + 10 = 30
Fourth term = Third term + d
= 30 + 10 = 40
Thus, the required first four terms of the AP are 10, 20, 30, 40.

2. a = -2, d = 0
First term = a = -2
Second term = First term + d
= -2 + 0 = -2
Third term = Second term + d
= -2 + 0 = -2
Fourth term = Third term + d
= -2 + 0 = -2
Thus, the required first four terms of the AP are -2, -2, -2, -2.

3. a = 4, d = -3
First term = a = 4
Second term = First term + d
= 4 + (-3) = 1
Third term = Second term + d
= 1 + (-3) = -2
Fourth term = Third term + d
= (-2) + (-3) = -5
Thus, the required first four terms of the AP are 4, 1, -2, -5.

4. a = -1, d = \(\frac{1}{2}\)
First term = a = -1
Second term = First term + d
= -1 + \(\frac{1}{2}\) = –\(\frac{1}{2}\)
Third term = Second term + d
= \(-\frac{1}{2}+\frac{1}{2}\) = 0
Fourth term = Third term + d
= 0 + \(\frac{1}{2}\) = \(\frac{1}{2}\)
Thus, the required first four terms of the AP are -1, – \(\frac{1}{2}\), 0, \(\frac{1}{2}\).

5. a = -1.25, d = -0.25
The general form of an AP is a, a + d, a + 2d, a + 3d, …….. Then,
First term = a = -1.25
Second term = a + d
= -1.25 + (-0.25) = -1.50
Third term = a + 2d
= -1.25 + 2(-0.25) = -1.75
Fourth term = a + 3d
= -1.25 + 3(-0.25) = -2.00
Thus, the required first four terms of the AP are -1.25, -1.50, -1.75, -2.00.

Question 3.
For the following APs, write the first term and the common difference:
1. 3, 1, 1, -3, ……..
2. -5, -1, 3, 7, …….
3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
4. 0.6, 1.7, 2.8, 3.9, …
Solution:
1. 3, 1, -1, -3,….
First term a = 3
Common difference d = a₂ – a₁ = 1 – 3 = -2

2. -5, 1, 3, 7,…..
First term a = -5
Common difference d = (-1) – (-5) = 4

3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
First term a = \(\frac{1}{3}\)
Common difference d = \(\frac{5}{3}-\frac{1}{3}=\frac{4}{3}\)

4. 0.6, 1.7, 2.8, 3.9, ……
First term a = 0.6
Common difference d = 1.7 – 0.6 = 1.1

Question 4.
Which of the following are APs? If they form an AP find the common difference d and write three more terms:
1. 2, 4, 8, 16, …….
2. 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …….
3. -1.2, -3.2, -5.2, -7.2, …….
4. -10, -6, -2, 2, …….
5. 3, 3 + 2\(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\), ….
6. 0.2, 0.22, 0.222, 0.2222, ……..
7. 0, -4, -8, -12, ……..
8. \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \cdots\)
9. 1, 3, 9, 27, ……
10. a, 2a, 3a, 4a, …….
11. a, a², a³, a⁴, …….
12. \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\), ……
13. \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\), …..
14. 1², 3², 5², 7², …….
15. 1², 5², 7², 7², …….
Solution:
2, 4, 8, 16, ….
a₂ – a₁ = 4 – 2 = 2,
a₃ – a₂ = 8 – 4 = 4,
a₄ – a₃ = 16 – 8 = 8
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

2. 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …
a₂ – a₁ = \(\frac{5}{2}-2=\frac{1}{2}\)
a₃ – a₂ = \(3-\frac{5}{2}=\frac{1}{2}\)
a₄ – a₃ = \(\frac{7}{2}-3=\frac{1}{2}\)
Here, a_k+1 – a_k is the same everywhere.
So, the given list of numbers forms an AP with d = \(\frac{1}{2}\)
Next three terms are given by-
a₅ = \(\frac{7}{2}+\frac{1}{2}=4\),
a₆ = \(4+\frac{1}{2}=\frac{9}{2}\) and
a₇ = \(\frac{9}{2}+\frac{1}{2}=5\)

3. -1.2, -3.2, -5.2, -7.2, …..
a₂ – a₁ = -3.2 – (-1.2) = -2
a₃ – a₂ = -5.2 – (-3.2) = -2
a₄ – a₃ = -7.2 – (-5.2) = -2
Here, a_k+1 – a_k is the same everywhere. So, the given list of numbers forms an AP with d = -2.
Next three terms are given by-
a₅ = -7.2 + (-2) = -9.2.
a₆ = -9.2 + (-2) = -11.2 and
a₇ = -11.2 + (-2) = -13.2

4. -10, -6, -2, 2, ……..
a₂ – a₁ = (-6) – (-10) = 4.
a₃ – a₂ = (-2) – (-6) = 4.
a₄ – a₃ = 2 – (-2) = 4
Here, a_k+1 – a_k is the same everywhere.
So, the given list of numbers forms an AP with d = 4.
Next three terms are given by-
a₅ = 2 + 4 = 6,
a₆ = 6 + 4 = 10 and
a₇ = 10 + 4 = 14

5. 3, 3+\(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\), …
a₂ – a₁ = 3 + \(\sqrt{2}\) – 3 = \(\sqrt{2}\),
a₃ – a₂ = (3 + 2\(\sqrt{2}\)) – (3 + \(\sqrt{2}\)) = \(\sqrt{2}\)
a₄ – a₃ = (3 + 3\(\sqrt{2}\)) – (3 + 2\(\sqrt{2}\)) = \(\sqrt{2}\)
Here, a_k+1 – a_k is the same everywhere.
So, the given list of numbers forms an AP with d = \(\sqrt{2}\).
Next three terms are given by-
a₅ = (3 + 3\(\sqrt{2}\)) + \(\sqrt{2}\) = 3 + 4\(\sqrt{2}\),
a₆ = (3 + 4\(\sqrt{2}\)) + \(\sqrt{2}\) = 3 + 5\(\sqrt{2}\) and
a₇ = (3 + 5\(\sqrt{2}\)) + \(\sqrt{2}\) = 3 + 6\(\sqrt{2}\)

6. 0.2, 0.22, 0.222. 0.2222,…
a₂ – a₁ = 0.22 – 0.2 = 0.02,
a₃ – a₂ = 0.222 – 0.22 = 0.002
Here, a₂ – a₁ ≠ a₃ – a₂.
Hence, the given list of numbers does not form an AP.

7. 0, -4, -8, -12, ….
a₂ – a₁ = (-4) – 0 = -4,
a₃ – a₂ = (-8) – (-4) = -4,
a₄ – a₃ = (-12) – (-8) = -4
Here, a_k+1 – a_k is the same everywhere.
Hence, the given list of numbers forms an AP with d = -4.
Next three terms are given by-
a₅ = (-12) + (-4) = -16,
a₆ = (-16) + (-4) = -20 and
a₇ = (-20) + (-4) = -24.

8.

Here, a_k+1 – a_k is the same everywhere. Hence, the given list of numbers forms an AP with d = 0.
Next three terms are given by-

9. 1, 3, 9, 27, ….
a₂ – a₁ = 3 – 1 = 2,
a₃ – a₂ = 9 – 3 = 6
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

10. a, 2a, 3a, 4a…
a₂ – a₁ = 2a – a = a,
a₃ – a₂ = 3a – 2a = a.
a₄ – a₃ = 4a – 3a = a
Here, a_k+1 – a_k is the same everywhere.
Hence, the given list of unknown numbers forms an AP with d = a.
Next three terms are given by-
a₅ = 4a + a = 5a,
a₆ = 5a + a = 6a and
a₇ = 6a + a = 7a

11. a, a², a³, a⁴,….
a₂ – a₁ = a² – a = a (a – 1),
a₃ – a₂ = a³ – a² = a² (a – 1)
Here, a₂ – a₁ ≠ a₃ – a₂.
Hence, the given list of unknown numbers does not form an AP.

12. \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\), ….
We know, \(\sqrt{8}\) = \(\sqrt{4 \times 2}\) = 2\(\sqrt{2}\).
\(\sqrt{18}\) = \(\sqrt{9 \times 2}\) = 3\(\sqrt{2}\) and
\(\sqrt{32}\) = \(\sqrt{16 \times 2}\) = 4\(\sqrt{2}\).
Hence, the given list of numbers is
\(\sqrt{2}\), 2\(\sqrt{2}\), 3\(\sqrt{2}\), 4\(\sqrt{2}\), ….
a₂ – a₁ = 2\(\sqrt{2}\) – \(\sqrt{2}\) = \(\sqrt{2}\),
a₃ – a₂= 3\(\sqrt{2}\) – 2\(\sqrt{2}\) = \(\sqrt{2}\)
a₄ – a₃ = 4\(\sqrt{2}\) – 3\(\sqrt{2}\) = \(\sqrt{2}\)
Here, a_k+1 – a_k is the same everywhere.
Hence, the given list of numbers forms an AP with d = \(\sqrt{2}\).
Next three terms are given by-
a₅ = 4\(\sqrt{2}\) + \(\sqrt{2}\) = 5\(\sqrt{2}\) = \(\sqrt{50}\).
a₆ = 5\(\sqrt{2}\) + \(\sqrt{2}\) = 6\(\sqrt{2}\) = \(\sqrt{72}\) and
a₇ = 6\(\sqrt{2}\) + \(\sqrt{2}\) = 7\(\sqrt{2}\) = \(\sqrt{98}\).

13. \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\), ……
a₂ – a₁ = \(\sqrt{6}\) – \(\sqrt{3}\) = \(\sqrt{3}\)(\(\sqrt{2}\) – 1)
a₃ – a₂ = \(\sqrt{9}\) – \(\sqrt{6}\) = \(\sqrt{3}\)(\(\sqrt{3}-\sqrt{2}\))
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

14. 1², 3², 5², 7², ….
a₂ – a₁ = 3² – 1² = 9 – 1 = 8,
a₃ – a₂ = 5² – 3² = 25 – 9 = 16
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

15. 1², 5², 7², 7²,…
a₂ – a₁ = 5² – 1² = 25 – 1 = 24,
a₃ – a₂ = 7² – 5² = 49 – 25 = 24,
a₄ – a₃ = 73 – 7² = 73 – 49 = 24.
Here. a_k+1 – a_k is the same everywhere.
Hence, the given list of numbers forms an AP with d = 24.
Next three terms are given by-
a₅ = 73 + 24 = 97,
a₆ = 97 + 24 = 121 and
a₇ = 121 + 24 = 145.