The ratio of the sum of n terms of two APs is (3n + 4): (2n + 7). Find the ratio of their mth terms.

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Solution

Let $a_{1}, a_{2}$ and $d_{1}, d_{2}$ be the first terms and common difference of two A.P’s
$S_{n_{1}}$ and $S_{n_{2}}$ denote the sum of n terms of two A.P’s
$\frac{S_{n_{1}}}{S_{n_{2}}} = \frac{7 n + 1}{4 n + 27}$ (Given)
$∴ \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{7 n + 1}{4 n + 27}$
$\Rightarrow \frac{[2 a_{1} + (n - 1) d_{1}]}{[2 a_{2} + (n - 1) d_{2}]} = \frac{7 n + 1}{4 n + 27}$
$\Rightarrow \frac{a_{1} + (\frac{n - 1}{2}) d_{1}}{a_{2} + (\frac{n - 1}{2}) d_{2}} = \frac{7 n + 1}{4 n + 27}$
Putting $\frac{n - 1}{2} = m - 1$ , we get
$\frac{a_{1} + (m - 1) d_{1}}{a_{2} + (m - 1) d_{2}} = \frac{7 \times (2 m - 1) + 1}{4 \times (2 m - 1) + 27} (∵ \frac{n - 1}{2} = m - 1 \Rightarrow n - 1 = 2 m - 2 \Rightarrow n = 2 m - 1)$
$\Rightarrow \frac{a_{m_{1}}}{a_{m_{2}}} = \frac{14 m - 6}{8 m + 23}$
$\Rightarrow a_{m_{1}} : a_{m_{2}} = 14 m - 6 : 8 m + 23$
Thus, the ratio of their $m^{t h}$ terms is $14 m - 6 : 8 m + 23$ .

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