If the ratio of the sum of the first n terms of two $A . P . s$ is $(7 n + 1) : (4 n + 27)$ , then find the ratio of their $9^{t h}$ terms.

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Solution

Given: The ratio of sum of $n$ terms of two AP's $= (7 n + 1) : (4 n + 27)$

Let the sum of n terms of first A.P and second A.P be $S_{n 1}$ and $S_{n 2}$ respectively.

So, $\frac{S_{n 1}}{S_{n 2}} = \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]}$

$\frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{7 n + 1}{4 n + 27}$

$\frac{a_{1} + \frac{n - 1}{2} d_{1}}{a_{2} + \frac{n - 1}{2} d_{2}} = \frac{7 n + 1}{4 n + 27}$ ---- (1)

Assume, $\frac{n - 1}{2} = m - 1$

$n = 2 m - 1$

From (1), we get

$\frac{a_{1} + (m - 1) d_{1}}{a_{2} + (m - 1) d_{2}} = \frac{7 (2 m - 1) + 1}{4 (2 m - 1) + 27}$

$\frac{a_{m 1}}{a_{m 2}} = \frac{14 m - 6}{8 m + 23}$

Thus the ratio of $m^{t h}$ terms of two AP’s is $(14 m - 6) : (8 m + 23) .$

Thus the ratio of $9^{t h}$ terms of two AP’s is $24 : 19$

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