If the sums of $n$ terms of two arithmetic progressions are in the ratio $2 n + 5 : 3 n + 4$ , then write the ratio of their $m^{t h}$ terms.

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Solution

It is given that

$\frac{S_{n}}{S_{n}^{1}} = \frac{2 n + 5}{3 n + 4}$

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

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

$∴ \frac{S_{n}}{S_{n}^{^{'}}} = \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{2 n + 5}{3 n + 4}$

$\frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{2 n + 5}{3 n + 4}$

$\frac{a_{1} + (\frac{n - 1}{2}) d_{1}}{a_{2} + (\frac{n - 1}{2}) d_{2}} = \frac{2 n + 5}{3 n + 4}$

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

$\Rightarrow n - 1 = 2 m$

$\Rightarrow n = 2 m - 1$

Replacing $\frac{n - 1}{2}$ by m on both side of equation (ii), we get

$\frac{a_{1} + m d_{1}}{a_{2} + m d_{2}} = \frac{2 (2 m - 1) + 5}{3 (2 m - 1) + 4}$

$= \frac{4 m - 2 + 5}{6 m - 3 + 4}$

$= \frac{4 m + 3}{6 m + 1}$

$∴$ Ratio of the mth terms $= \frac{4 m + 3}{6 m + 1}$

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