If the ratio of the sum of first n terms of two A.Ps in $(7 n + 1) : (4 n + 27)$ , then find ratio of their $9^{t h}$ terms.

24 : 90

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24 : 29

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24 : 19

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34 : 19

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Solution

The correct option is B $24 : 19$
$A P 1 \to a_{1} . . . . . . . a_{n}$ common diff. $= d_{1}$
$A P 2 \to b_{1} . . . . . . . . b_{n}$ common diff $= d_{2}$
Given ratio of sum n terms of $2 A P$ is $(7 n + 1) : (4 n + 27)$
$\frac{\frac{n}{2} (2 b_{1} + (n - 1) d_{1})}{\frac{n}{2} (2 b_{1} + (n - 1) d_{2})} = \frac{7 n + 1}{4 n + 27}$

$\frac{d_{1} n + (2 a_{1} - d_{1})}{d_{2} n + (2 b_{2} - d_{2})} = \frac{7 n + 1}{4 n + 27} \to d_{1} = 7 x, d_{2} = 4 x$

$a_{1} = 4 x, b_{1} = 15.5 x$

then ratio of $9^{t h}$ term = $\frac{a_{1} + 8 d_{1}}{b_{1} + 8 d_{2}} = \frac{4 x + 56 x}{15.5 x + 32 x}$

$= \frac{60 x}{47.5 x} = 120 : 95$

$= 24 : 19$

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If ratio of the sum of first n terms of two A.P is 7n+1:4n+27,then find the ratio of their 9th terms.

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