Question

The Sum of first n terms of two Aps are in ratio
(3n+8): (7n+15), find the ratio of their 12th term.
​

Answer 1

Step-by-step explanation:

Used formula: Sn = n/2 [2a + (n - 1)d]

As per given condition,

First term:

→ (a + (n - 1)d)/(a + (n - 1)d)

→ (a + (12 - 1)d)/(a + (12 - 1)d)

→ (a + 11d)/(a + 11d)

Sum of n of terms:

→ [2a + (n - 1)d]/[(2a + (n - 1)d] = (3n + 8)/(7n + 15)

→ [a + (n - 1)/2)d]/[a + (n - 1)/2)d] = (3n + 8)/(7n + 15) ......(1)

So,

→ (n - 1)/2 = 11

→ n - 1 = 22

→ n = 23

Substitute value of n in (1)

→ [a + (n - 1)/2)d]/[a + (n - 1)/2)d] = (3*23 + 8)/(7*23 + 15)

→ [a + (n - 1)/2)d]/[a + (n - 1)/2)d] = (69 + 8)/(161 + 15)

→ [a + (n - 1)/2)d]/[a + (n - 1)/2)d] = 77/176

→ [a + (n - 1)/2)d]/[a + (n - 1)/2)d] = 7/16

Hence, the ratio of their 12th term is 7:16.