If the ratio of sum of $n$ terms in two A.P's is $2 n : n + 1$ , then the ratio of $8^{t h}$ terms is

15 : 8

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8 : 133

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5 : 17

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none

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Solution

The correct option is B $15 : 8$
Let the first term and common difference of the first AP be $a_{1}$ and $d_{1}$ and the same for the second AP be $a_{2}$ and $d_{2}$
So,
$\frac{2 n}{n + 1} = \frac{\frac{n}{2} (2 a_{1} + (n - 1) d_{1})}{\frac{n}{2} (2 a_{2} + (n - 1) d_{2})}$
$\Rightarrow \frac{2 n}{n + 1} = \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}}$
$\Rightarrow \frac{2 n}{n + 1} = \frac{a_{1} + (\frac{n - 1}{2}) d_{1}}{a_{2} + (\frac{n - 1}{2}) d_{2}}$
$\Rightarrow \frac{2 n}{n + 1} = \frac{a_{1} + N d_{1}}{a_{2} + N d_{2}}$
Where $N = \frac{n - 1}{2}$
For the ratio of $8^{t h}$ terms $N = 7$
Or
$\frac{n - 1}{2} = 7$
$\Rightarrow n - 1 = 14$ $\Rightarrow n = 15$
Thus,
$\frac{a_{1} + 7 d_{1}}{a_{2} + 7 d_{2}}$ $= \frac{2 (15)}{15 + 1}$
$\Rightarrow \frac{2 (15)}{16}$ $= 15 : 8$

$∴$ the ratio of 8th term of the first and second AP is $15 : 8$

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