The sum of n terms of two series in A.P. are in the ratio $(3 n - 13) (5 n + 21)$ . Find the ratio of their $24^{t h}$ terms.

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Solution

$L e t sin = \frac{n}{2} [2 a + (n - 1) D] e q n . (i)$
$a n d sin = \frac{n}{2} [2 A + (n - 1) D] e q n . (i i)$
$T h e i r r a t i o i s (3 n - 13) : (5 n + 2 i)$
$\Rightarrow \frac{s n}{s^{'} n} = \frac{[2 a + (n - 1) d]}{[2 A + (n - 1) D]} = \frac{(3 n - 1)}{(5 x + 21)} e q n . (i i i)$
$∴ R a t i o o f 24^{t h} t e r m$
$\Rightarrow \frac{T_{2 y}}{t_{2 y}} = \frac{a + 23 d}{A + 23 D} = \frac{2 a + 46 d}{2 A + 46 D} = \frac{2 a + (45 - 1) d}{2 A + (45 - 1 d}$
$P u t t i n g n = 45 i n e q n . (i i i) \frac{3 \times 45 - 13}{5 \times 45 + 21} = \frac{135 - 13}{2^{o} 25 + 21} = \frac{122}{246} = \frac{61}{123}$

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