If the ratio of sum of first $n$ terms of two A.P.s is $2 n + 8 : 5 n - 3,$ then the ratio of $n^{th}$ terms of those two A.P.s is

2 n + 5 : 5 n + 1

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2 n + 3 : 5 n - 4

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2 n + 18 : 5 n - 1

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2 n + 18 : 5 n + 1

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Solution

The correct option is B $2 n + 3 : 5 n - 4$
Let, for $1^{st}$ A.P., first term $= a_{1}$ and common difference $= d_{1}$
and, for $2^{nd}$ A.P., first term $= a_{2}$ and common difference $= d_{2}$
$\frac{(S_{n})_{1}}{(S_{n})_{2}} = \frac{2 n + 8}{5 n - 3}$
$\Rightarrow \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{2 n + 8}{5 n - 3}$
$\Rightarrow \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{2 n + 8}{5 n - 3}$

Replacing $n$ with $(2 n - 1)$ in above equation, we get
$\frac{2 a_{1} + (2 n - 2) d_{1}}{2 a_{2} + (2 n - 2) d_{2}} = \frac{4 n + 6}{10 n - 8}$
$\Rightarrow \frac{a_{1} + (n - 1) d_{1}}{a_{2} + (n - 1) d_{2}} = \frac{2 n + 3}{5 n - 4}$
$∴ \frac{(t_{n})_{1}}{(t_{n})_{2}} = \frac{2 n + 3}{5 n - 4}$

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