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

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Solution

Ratio of sum of first n terms of two A.P.s are

$\frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{n}{2} [2 A + (n - 1) D]} = \frac{7 n + 1}{4 n + 27}$

Put n = 17

$\Rightarrow \frac{2 a + (16) d}{2 A + (16) D} = \frac{120}{95}$

$\frac{2 a + (16) d}{2 A + (16) D} = \frac{120}{95} = \frac{24}{19}$

$\frac{a + 8 d}{A + 8 D} = \frac{24}{19}$
$∴$ $\frac{a_{9}}{A_{9}} = \frac{24}{19}$

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