Class 10 Maths Chapter 5 Question Answers - Arithmetic Progressions - 2

Q17. If S_n the sum of n terms of an A.P. is given by S_n = 3n² - 4n, find the nth term.
Sol. We have:
S_{n - 1} = 3 (n - 1)² - 4 (n - 1)
= 3 (n² - 2n + 1) - 4n + 4
= 3n² - 6n + 3 - 4n + 4
= 3n² - 10n + 7
∵ nth term = S_n - S_{n - 1}
= 3n² - 4n - [3n² - 10n + 7]
= 3n² - 4n - 3n² + 10n - 7
= 6n - 7.

Q18. The sum of 4th and 8th terms of an A.P. is 24, and the sum of 6th and 10th terms is 44. Find the A.P.
Sol. Let, the first term = a
Common difference be = d
∴ Using T_n = a + (n - 1) d, we have
T₄ = a + 3d
T₆ = a + 5d
T₈ = a + 7d
T₁₀ = a + 9d
∵ T₄ + T₈ = 24
∴ (a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12
[Dividing by 2]    ...(1)
Also T₆ + T₁₀ = 44
∴ (a + 5d) + (a + 9d) = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22
[Dividing by 2]   ...(2)
Subtracting (1) from (2), we have:
(a + 7d) - (a + 5d) = 22 - 12
⇒ 2d = 10 ⇒ d = 5
From (1), a + 5 (5) = 12
⇒ a = 12 - 25 = - 13
Since, the A.P. is given by:
a, a + d, a + 2d, .....
∴ We have the required A.P. as:
- 13, (- 13 + 5), [- 13 + 2 (5)], .....
or - 13, - 8, - 3, .....

Q19. If Sn, the sum of first n terms of an A.P. is given by 
S_n = 5n² + 3n
Then find the nth term.
 Sol. ∵ S_n = 5n² + 3n
∴ S_{n - 1} = 5 (n - 1)² + 3 (n - 1)
= 5 (n² - 2n + 1) + 3 (n - 1)
= 5n² - 10n + 5 + 3n - 3
= 5n² - 7n + 2
Now, nth term = S_n - S_{n - 1}
∴ The required nth term
= [5n² + 3n] - [5n² - 7n + 2]
= 10n - 2.

Q20. The sum of 5th and 9th terms of an A.P. is 72 and the sum of 7th and 12th term of 97. Find the A.P. 
Sol. Let ‘a’ be the 1st term and ‘d’ be the common difference of the A.P.
Now, using Tn = a + (n - 1) d, we have
T₅ = a + 4d
T₇ = a + 6d
T₉ = a + 8d
T₁₂ = a + 11d
∵ T₅ + T₉ = 72
∴ a + 4d + a + 8d = 72
⇒ 2a + 12d = 72
⇒ a + 6d = 36
[Dividing by 2]   ...(1)
Also T₇ + T₁₂= a + 6d + a + 11d = 97
⇒ 36 + a + 11d = 97 [From (1)]
⇒ a + 11d = 97 - 36
⇒ a + 11d = 61    ...(2)
Subtracting (1) from (2), we get
a + 11d - a - 6d = 61 - 36
⇒ 5d = 25
⇒ d = 25/5
From (1), we have
a + 11 (5) = 61
a + 55 = 61
⇒ a = 61 - 55 = 6
Now, a_n A.P. is given by
a, a + d, a + 2d, a + 3d, .....
∴ The required A.P. is:
6, (6 + 5), [6 + 2 (5)], [6 + 3 (5)], .....
or 6, 11, 16, 24, .....

Q21. In an A.P. the sum of its first ten terms is –150 and the sum of its next term is –550. Find the A.P.
Sol. Let the first term = a
And the common difference = d
∴
⇒ 10a + 45d = –150
⇒ 2a + 9d = –30   ...(1)
∵ The sum of next 10 terms
(i.e. S₂₀ – S₁₀) = –550

⇒ 20a + 190d + 150 = –550
⇒ 2a + 19d + 15 = –55
⇒ 2a + 19d = – 55 – 15
⇒ 2a + 19d = –70   ...(2)
Subtracting (1) from (2), we get

From (1), 2(a) + 9(–4) = –30 or a = 6/2 = 3
Thus, AP is a, a + d, a + 2d ...
or 3, [3 + (–4)], [3 + 2(–4)], ...
or 3, –1, –5, ...

Q22. Which term of the A.P. 3, 15, 27, 39, ..... will be 120 more than its 21st term?
Sol. Let the 1st term is ‘a’ and common difference = d
∴ a = 3 and d = 15 - 3 = 12
Now, using T_n = a + (n - 1) d
∴ T₂₁ = 3 + (21 - 1) × 12
= 3 + 20 × 12
= 3 + 240 = 243
Let the required term be the nth term.
∵ nth term = 120 + 21st term
= 120 + 243 = 363
Now T_n = a + (n - 1) d
⇒ 363 = 3 + (n - 1) × 12
⇒ 363 - 3 = (n - 1) × 12
⇒ n - 1 = 360/12 = 30
⇒ n = 30 + 1 = 31
Thus the required term is the 31st term of the A.P.

Q23. Which term of the A.P. 4, 12, 20, 28, ..... will be 120 more than its 21st term?
Sol. Here, a = 4
d = 12 - 4 = 8
Using T_n = a + (n - 1) d
∴ T₂₁ = 4 + (21 - 1) × 8
= 4 + 20 × 8 = 164
∵ The required nth term = T₂₁ + 120
∴ nth term = 164 + 120 = 284
∴ 284 = a + (n - 1) d
⇒ 284 = 4 + (n - 1) × 8
⇒ 284 - 4 = (n - 1) × 8
⇒ n - 1 = 280/8 = 35
⇒ n = 35 + 1 = 36
Thus, the required term is the 36th term of the A.P.

Q24. The sum of n terms of an A.P. is 5n² - 3n. Find the A.P. Hence find its 10th term.
Sol. We have:
S_n = 5n² - 3n
∴ S₁ = 5 (1)² - 3 (1) = 2
⇒ First term T1 = (a) = 2
S₂ = 5 (2)² - 3 (2)
= 20 - 6 = 14
⇒ Second term T₂ = 14 - 2 = 12
Now the common difference = T₂ - T₁
⇒ d = 12 - 2 = 10
∵ An A.P. is given by
a, (a + d), (a + 2d) .....
∴ The required A.P. is:
2, (2 + 10), [2 + 2 (10)], .....
⇒ 2, 12, 22, .....
Now, using T_n = a + (n - 1) d, we have
T₁₀ = 2 + (10 - 1) × 10
= 2 + 9 × 10
= 2 + 90 = 92.

Q25. Find the 10th term from the end of the A.P.:
8, 10, 12, ....., 126
Sol. Here, a = 8
d = 10 - 8 = 2
T_n = 126
Using T_n = a + (n - 1) d
⇒ 126 = 8 + (n - 1) × 2
⇒ n - 1 =
⇒ n = 59 + 1 = 60
∴ l = 60
Now 10th term from the end is given by
l - (10 - 1) = 60 - 9 = 51
Now, T₅₁ = a + 50d
= 8 + 50 × 2
= 8 + 100 = 108
Thus, the 10th term from the end is 108.

Q26. The sum of n terms of an A.P. is 3n² + 5n. Find the A.P. Hence, find its 16th term.
Sol. We have,
S_n = 3n² + 5n
∴ S₁ = 3 (1)² + 5 (1)
= 3 + 5 = 8
⇒ T₁ = 8 ⇒ a = 8
S₂ = 3 (2)² + 5 (2)
= 12 + 10 = 22
⇒ T₂ = 22 - 8 = 14
Now d = T₂ - T₁ = 14 - 8 = 6
∵ An A.P. is given by,
a, (a + d), (a + 2d),  .....
∴ The required A.P. is:
8, (8 + 6), [8 + 2 (6)],     .....
⇒ 8, 14, 20, .....
Now, using T_n = a + (n - 1) d, we hve
T₁₆ = a + 15d
= 8 + 15 × 6 = 98
Thus, the 16th term of the A.P. is 98.

Q27. In an AP, the sum of first n-terms is . Find the 25^th term.
Sol. We know that: an = S_n - S_n-1, where,

∴

Now, a₂₅ = S₂₅ − S₂₄

 
⇒
⇒
⇒

Q28. The sum of 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the first three terms of the A.P.
Sol. Let the first term be ‘a’ and the common difference be ‘d’.
Using T_n = a + (n - 1) d, we have
T₄ = a + 3d, T₆ = a + 5d
T₈ = a + 7d and T₁₀ = a + 9d
Since T₄ + T₈ = 24
∴ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24 ⇒ a + 5d = 12   ...(1)
Also, T₆ + T₁₀ = 44
∴  a + 5d + a + 9d = 44
⇒ 2a + 14d = 44 ⇒ a + 7d = 22   ...(2)
Subtracting (2) from (1), we get,
a + 7d - a - 5d = 22 - 12
⇒ 2d = 10 ⇒ d = 5
Now from (1),
a + 5 (5) = 12
⇒ a + 25 = 12 ⇒ a = - 13
∴ First term (T₁) = a + 0 = - 13
Second term (T₂) = a + d
= - 13 + 5 = - 8
Third term T₃ = - a + 2d
= - 13 + 10 = - 3

Q29. In an A.P., the first term is 8, nth term is 33 and sum of first n terms is 123. Find n and d, the common difference.
Sol. Here,
First term T₁ = 8 ⇒ a = 8
nth term T_n = 33 = l
∵ S_n = 123 [Given]
∴ Using, S_n = n/2 [a + l], we have
S_n = n/2 [8 + 33]
⇒
⇒
Now, T₆ = 33
⇒ a + 5d = 33
⇒ 8 + 5d = 33
⇒ 5d = 33 - 8 = 25
⇒ d = 25/5 = 5
Thus, n = 6 and d = 5.

Q30. For what value of n are the nth terms of two A.P.’s 63, 65, 67, ..... and 3, 10, 17, ..... equal?
Sol. For the 1st A.P.
a = 63
d = 65 - 63 = 2
∴ T_n = a + (n - 1) d
⇒ T_n = 63 + (n - 1) × 2
For the 2nd A.P.
a = 3
d = 10 - 3 = 7
∴ T_n = a + (n - 1) d
⇒ T_n = 3 + (n - 1) × 7
∵ [T_n of 1st A.P.] = [T_n of 2nd A.P.]
∴ 63 + (n - 1) × 2 = 3 + (n - 1) × 7
⇒ 63 - 3 + (n - 1) × 2 = (n - 1) 7
⇒ 60 + (n - 1) × 2 - (n - 1) × 7 = 0
⇒ 60 + (n - 1) [2 - 7] = 0
⇒ 60 + (n - 1) × (- 5) = 0
⇒ (n - 1) = -60/-5 = 12
⇒ n = 12 + 1 = 13
Thus, the required value of n is 13.

Q31. If m times the mth term of an A.P. is equal to n times the nth term, find the (m + n)th term of the A.P.
Sol. Let the first term (T₁) = a and the common difference be ‘d’.
∴ nth term = a + (n - 1) d
And mth term = a + (m - 1) d
Also,
(m + n)th term = a + (m + n - 1) d   ...(1)
∵ m (mth term) = n (nth term)
∴ m [a + (m - 1) d] = n [a + (n - 1) d]
⇒ ma + m (m - 1) d = na + n (n - 1) d
⇒ ma + (m² - m) d - na - (n² - n) d = 0
⇒ ma - na + (m² - m) d - (n² - n) d = 0
⇒ a [m - n] + [m² - m - n² + n] d = 0
⇒ a [m - n] + [(m² - n²) - (m - n)] d = 0
⇒ a [m - n] + [(m + n) (m - n) - (m - n)] d = 0
⇒ a [m - n] + (m - n) [m + n - 1] d = 0
Dividing throughout by (m - n), we have:
a + [m + n - 1] d = 0
⇒ a + [(m + n) - 1] d = 0     ...(2)
⇒ (m + n) th term = 0 [From (1) and (2)]

Q32. In an A.P., the first term is 25, nth term is - 17 and sum of first n terms is 60. Find ‘n’ and ‘d’, the common difference.
Sol. Here, the first term a = 25
And the nth term = - 17 = l
Using T_n = a + (n - 1) d, we have:
- 17 = 25 + (n - 1) d
⇒ (n - 1) d = - 17 - 25 = - 42

Also, S_n = n/2 [a + l]
⇒ 60 = n/2 [25 + (- 17)]
⇒
⇒ 60 = 4n ⇒ n = 60/4 = 15
From (1), we have

Thus, n = 15 and d = - 3

Q33. In an A.P., the first term is 22, nth term is - 11 and sum of first n terms is 66. Find n and d, the common difference. 
Sol. We have
1st term (T₁) = 22 ⇒ a = 22
Last term (T_n) = - 11 ⇒ l = - 11
Using, S_n = n/2 [a + l], we have:
66 = n/2 [22 + (- 11)]
⇒ 66 × 2 = n [11]
⇒
Again using
T_n = a + (n - 1) d
We have:
T₁₂ = 22 + (12 - 1) × d
- 11 = 22 + 11d [∵ nth term = - 11]
⇒ 11d = - 22 - 11 = - 33
⇒ d = -33/11 = -3
Thus, n = 12 and d = - 3