If the ratio between the sum of first n terms of two A.P's is $7 n + 1 : 4 n + 27$ then the ratio of $11$ th terms is :

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Solution

Let $a_{1}, d_{1}$ be the first term and common difference of $1^{s t}$ AP.
and $a_{2}, d_{2}$ be the first term and common difference of $2^{n d}$ AP.
Given that,
$\frac{\frac{n}{2} (2 a_{1} + (n - 1) d_{1})}{\frac{m}{2} (2 a_{2} + (n - 1) d_{2})} = \frac{7 n + 1}{4 n + 27}$
$\Rightarrow \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{7 n + 1}{4 n + 27}$ _________ (I)
Ratio of $11^{t h}$ terms of two AP's is
$\frac{a_{1} + (11 - 1) d_{1}}{a_{2} + (11 - 1) d_{2}} = \frac{a_{1} + 10 d_{1}}{a_{2} + 10 d_{2}} = \frac{2 (a_{1} + 10 d_{1})}{2 (a_{2} + 10 d_{2})}$
$\Rightarrow \frac{a_{1} + (11 - 1) d_{1}}{a_{2} + (11 - 1) d_{2}} = \frac{2 a_{1} + 20 d_{1}}{2 a_{1} + 20 d_{2}}$ _______ (II)
Substituting, m=21, in (I) we get
$\frac{2 a_{1} + (21 - 1) d_{1}}{2 a_{2} + 20 d_{2}} = \frac{7 (21) + 1}{4 (21) + 27}$
$\frac{2 a_{1} + 20 d_{1}}{2 a_{1} + 20 d_{2}} = \frac{148}{111}$
$\Rightarrow \frac{2 (a_{1} + 10 d_{1})}{2 (a_{2} + 10 d_{2})} = \frac{148}{111}$
$\Rightarrow \frac{a_{1} + 10 d_{1}}{a_{2} + 10 d_{2}} = \frac{4}{3}$
$∴$ Ratio of 11 terms is 483.

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