The sums of first n terms of two A.P’s are in the ratio (7n + 2) : (n + 4). Find the ratio of their 5^th terms.

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Solution

Let the first term, the common difference and the sum of the first A.P. be $a_{1}, d_{1}$ and $S_{1}$ , respectively, and those of the second A.P. be $a_{2}, d_{2}$ and $S_{2}$ , respectively.
Then, we have,
$S_{1} = \frac{n}{2} [2 a_{1} + (n - 1) d_{1}]$
And, $S_{2} = \frac{n}{2} [2 a_{2} + (n - 1) d_{2}]$
Given:
$\frac{s_{1}}{s_{2}} = \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{7 n + 2}{n + 4} \Rightarrow \frac{s_{1}}{s_{2}} = \frac{[2 a_{1} + (n - 1) d_{1}]}{[2 a_{2} (n - 1) d_{2}]} = \frac{7 n + 2}{n + 4}$
To find the ratio of the 5th terms of the two A.P.s, we replace n by
$2 \times 5 - 1 = 9$ in the above equation:
$\Rightarrow \frac{[2 a_{1} + (9 - 1) d_{1}]}{[2 a_{2} (9 - 1) d_{2}] = \frac{7 \times 9 + 2}{9 + 4}} \Rightarrow \frac{[2 a_{1} + (8) d_{1}]}{[2 a_{2} + (8) d_{2}]} = \frac{7 \times 9 + 2}{9 + 4} = \frac{65}{13} \frac{[a_{1} + 4 d_{1}]}{a_{2} + 4 d_{2}} = \frac{5}{1} = 5 : 1$

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