The sum of 'n' term of two arithmetic series are in ration of (7n + 1) : (4n + 27). Find the ratio of their $n^{t h}$ term.

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Solution

$\frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{7 n + 1}{4 n + 27}$
$\Rightarrow \frac{a_{1} (\frac{n - 1}{2}) d_{1}}{a_{2} (\frac{n - 1}{2}) d_{2}} = \frac{7 n + 1}{4 n + 27}$
$\frac{a_{1} (\frac{2 n - 1 - 1}{2}) d_{1}}{a_{2} (\frac{2 n - 1 - 1}{2}) d_{2}} = \frac{7 (2 n - 1) + 1}{4 (2 n - 1) + 27}$
$\Rightarrow \frac{[a_{1} + (n - 1) d_{1}]}{[a_{2} + (n - 1) d_{2}]} = \frac{14 n - 7 + 1}{8 n - 4 + 27} \Rightarrow \frac{{(T_{n})}_{1}}{{(T_{n})}_{2}} - \frac{14 n - 6}{8 n - 23}$

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