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

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Solution

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

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