The sum of $n$ terms of two $A . P . s$ are in the ratio $(3 n + 8) : (7 n + 15)$ .Determine the ratio of their $10 t h$ terms.

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Solution

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

Given,

$\frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{n}{2} [2 A + (n - 1) D]} = \frac{3 n + 8}{7 n + 15}$

$\frac{[2 a + (n - 1) d]}{[2 A + (n - 1) D]} = \frac{3 n + 8}{7 n + 15}$

$\frac{[a + \frac{(n - 1)}{2} d]}{[A + \frac{(n - 1)}{2} D]} = \frac{3 n + 8}{7 n + 15}$ ...(1)

we need to find 10th term, given by,

$= \frac{a + 9 d}{A + 9 D}$

$∴ \frac{(n - 1)}{2} = 9$

$n - 1 = 18$

$n = 19$

substitute n in eqn (1)

$\frac{[a + \frac{(19 - 1)}{2} d]}{[A + \frac{(19 - 1)}{2} D]} = \frac{3 (19) + 8}{7 (19) + 15}$

$\frac{[a + 9 d]}{[A + 9 D]} = \frac{65}{148}$

Hence the ratio of 10th term is $\frac{65}{148} \to 65 : 148$

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