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The sum of first n terms of two APs are in the ratio (3 n+8):(7 n+15). What will be the ratio of their 12th terms?

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6:15

8:17

9:16

7:16

answer is D.

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Detailed Solution

Given that the sum of first n terms of two APs are in the ratio

(3 n + 8) : 7 n + 15

.
Determining first A.P.:
Let the first term = a and common difference = d.
Applying formula for sum of first n terms of first A.P.:

∴ s n = n 2 [2 a + (n − 1) d]

Formula used for nth term of A.P.:

t n = a + n − 1 d

Applying formula for nth term of A.P.:

∴ t 12 = a + (12 − 1) d ⇒ t 12 = a + 11 d ∴ a + 11 d = 3 n + 8

Determining second A.P.:
Let the first term = A and common difference = D.
Applying formula for sum of first n terms of second A.P.:

∴ S n = n 2 [2 A + (n − 1) D]

Applying formula for nth term of A.P.:

∴ T 12 = A + (12 − 1) D ⇒ T 12 = A + 11 D ∴ A + 11 D = 7 n + 15

As for 12th term:
Question Image

⇒ a + n − 1 2 d A + n − 1 2 D = 3 n + 8 7 n + 15 ⇒ a + 11 d A + 11 D = 3 n + 8 7 n + 15 − − − − (1)

∴ n − 1 2 = 11 ⇒ n − 1 = 22 ⇒ n = 23

Putting the value of n = 23 in equation (1):

∴ 3 × 23 + 8 7 × 23 + 15 = a + 11 d A + 11 D ⇒ a + 11 d A + 11 D = 77176 = 716

Therefore, the ratio of 12th terms is 7:16
Hence, option (4) is correct.

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