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

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Solution

Let $a_{1}$ , $a_{2}$ be the first terms and $d_{1}$ , $d_{2}$ the common differences of the two given A.P.'s. Then, the sums of their n terms are given by
$S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d_{1}]$
And,
${S_{n}}^{'} = \frac{n}{2} [2 a_{2} + (n - 1) d_{2}]$

Therefore,
$\frac{S_{n}}{{S_{n}}^{'}} = \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}}$

It is given that
$\frac{S_{n}}{{S_{n}}^{'}} = \frac{7 n + 1}{4 n + 27}$

$⟹ \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{7 n + 1}{4 n + 27}$ ....(1)

To find the ratio of the mth terms of the two given AP's, we replace n by (2m-1) in equation 1.
Therefore,
$\frac{2 a_{1} + (2 m - 2) d_{1}}{2 a_{2} + (2 m - 2) d_{2}} = \frac{7 (2 m - 1) + 1}{4 (2 m - 1) + 27}$

$⟹ \frac{a_{1} + (m - 1) d_{1}}{a_{2} + (m - 1) d_{2}} = \frac{14 m - 6}{8 m + 23}$

Hence, the ratio of the mth terms of two AP's is $(14 m - 6) : (8 m + 23)$ .

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