The ratio of the sum of $n$ terms of two A.P's is $(7 n + 1) : (4 n + 27)$ . Find the ratio of their $m^{t h}$ terms.

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Solution

Let $a_{1}, a_{2}$ be the first terms and $d_{1}, d_{2}$ the common diference of the two given A.P's. Then the sums of their $n$ terms are given by
$S_{n} = \frac{n}{2} {2 a_{1} + (n - 1) d_{1}}$ , and $S_{n^{'}} = \frac{n}{2} {2 a_{2} + (n - 1) d_{2}}$
$∴ \frac{S_{n}}{S_{n^{'}}} = \frac{\frac{n}{2} {2 a_{1} + (n - 1) d_{1}}}{\frac{n}{2} {2 a_{2} + (n - 1) d_{2}}} = \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}}$
It is given that
$\frac{S_{n}}{S_{n^{'}}} = \frac{7 n + 1}{4 n + 27}$
$⟹ \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{7 n + 1}{4 n + 27}$ ...(i)
To find the ratio of the terms of the $m^{t h}$ terms of the two given A.P's, we replace $n$ by $(2 m - 1)$ in equation (i).
Replacing $n$ by $(2 m - 1)$ in equation (i), we get
$\frac{2 a_{1} + (2 m - 2) d_{1}}{2 a_{2} + (2 m - 2) d_{2}} = \frac{7 (2 m - 1) + 1}{4 (2 m - 1) + 27}$
$⟹ \frac{a_{1} + (m - 1) d_{1}}{a_{2} + (m - 1) d_{2}} = \frac{14 m - 6}{8 m + 23}$
Hence, the ratio of the $m^{t h}$ terms of the two A.P's is $(14 m - 6) : (8 m + 23)$ .

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