The sum of $n$ terms of two A.P.'s are in the ratio $(5 n + 21) : (3 n - 13)$ . Find the ratio of their $24^{t h}$ terms.

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Solution

The ratio of n terms of two AP's $= \frac{(5 n + 21)}{(3 n - 13)}$
The sum of an AP $= \frac{n}{2} (2 a + (n - 1) d)$

$\frac{\frac{n}{2} (2 a_{1} + (n - 1) d_{1})}{\frac{n}{2} (2 a_{2} + (n - 1) d_{2})} = \frac{(5 n + 21)}{(3 n - 13)}$

$\frac{(2 a_{1} + (n - 1) d_{1})}{(2 a_{2} + (n - 1) d_{2})} = \frac{(5 n + 21)}{(3 n - 13)}$

24th term of an AP $= a + (24 - 1) d$
$= a + 23 d$

put n=47

$\frac{(2 a_{1} + (47 - 1) d_{1})}{(2 a_{2} + (47 - 1) d_{2})} = \frac{(5 \times 47 + 21)}{(3 \times 47 - 13)}$

$\frac{(2 a_{1} + (46 d_{1})}{(2 a_{2} + (46) d_{2})} = \frac{(235 + 21)}{(141 - 13)}$

$\frac{a_{1} + (23) d_{1}}{a_{2} + (23) d_{2}} = \frac{256}{128}$

$\frac{a_{1} + (23) d_{1}}{a_{2} + (23) d_{2}} = \frac{2}{1}$
so the ratio of the 24th term $= 2 : 1$

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