Question

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

Answer 1

Given: Let the two APs have first terms a₁, a₂ and common differences d₁, d₂.

The sums of first n terms satisfy `(S1_n)/(S2_n) = (3n + 8) : (7n + 15)`, with `S_n = n/2 [2a + (n - 1)d]`.

Step-wise calculation:

1. Write ratio of sums using S_n formula:

`(n/2 (2a_1 + (n - 1)d_1))/(n/2 (2a_2 + (n - 1)d_2)) = (3n + 8)/(7n + 15)` 

⇒ `(2a_1 + (n - 1)d_1)/(2a_2 + (n - 1)d_2) = (3n + 8)/(7n + 15)`

2. To get the ratio of the mth terms, T_m = a + (m – 1)d.

Substitute n = 2m – 1 so that 2a + (n – 1)d = 2a + (2m – 2)d = 2[a + (m – 1)d]. 

Thus, `(a_1 + (m - 1)d_1)/(a_2 + (m - 1)d_2)` 

= `(3(2m - 1) + 8)/(7(2m - 1) + 15)` 

= `(6m + 5)/(14m + 8)`

3. For the 12th terms (m = 12):

Ratio = (6 × 12 + 5) : (14 × 12 + 8) = 77 : 176.

Simplify by gcd 11 → 7 : 16.

The ratio of their 12th terms is 7 : 16.