The sum of $n$ terms of two arithmetic progressions are in the ratio $(7 n + 1) : (4 n + 17)$ . Find the ratio of their $n$ th terms.

\frac{14 n - 6}{8 n - 23}

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\frac{14 n + 6}{8 n + 23}

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\frac{14 n + 6}{8 n - 23}

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\frac{14 n - 6}{8 n + 13}

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Solution

The correct option is D $\frac{14 n - 6}{8 n + 13}$
$\frac{\frac{n}{2} (2 a_{1} + (n - 1) d_{1})}{\frac{n}{2} (2 a_{2} + (n - 1) d_{2})} = \frac{7 n + 1}{4 n + 17}$
$\Rightarrow \frac{(2 a_{1} + (n - 1) d_{1})}{(2 a_{2} + (n - 1) d_{2})} = \frac{7 n + 1}{4 n + 17}$
Let $n - 1 = 2 m - 2$
So we get $n = 2 m - 1$
$\frac{(2 a_{1} + 2 (m - 1) d_{1})}{(2 a_{2} + 2 (m - 1) d_{2})} = \frac{7 (2 m - 1) + 1}{4 (2 m - 1) + 17} = \frac{14 m - 6}{8 m + 13}$
$\frac{(a_{1} + (m - 1) d_{1})}{(a_{2} + (m - 1) d_{2})} = \frac{14 m - 6}{8 m + 13}$

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