If the sums of n terms of two arithmetic progressions are in the ratio 2n + 5 : 3n + 4, then write the ratio of their mth terms.

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Solution

Given:

$\frac{S_{n}}{{S_{n}}^{1}} = \frac{2 n + 5}{3 n + 4} \Rightarrow \frac{\frac{n}{2} \{2 a + (n - 1) d\}}{\frac{n}{2} \{2 b + (n - 1) d^{1}\}} = \frac{2 n + 5}{3 n + 4} \Rightarrow \frac{2 a + (n - 1) d}{2 b + (n - 1) d^{1}} = \frac{2 n + 5}{3 n + 4} . . . (1)$

Ratio of their m terms = $\frac{a_{m}}{b_{m}}$
To find the ratio of the m^th terms, replace n by 2m $-$ 1 in equation (1).
$\Rightarrow \frac{2 a + (2 m - 2) d}{2 b + (2 m - 2) d^{1}} = \frac{2 (2 m - 1) + 5}{3 (2 m - 1) + 4} \Rightarrow \frac{a + (m - 1) d}{b + (m - 1) d^{1}} = \frac{4 m - 2 + 3}{6 m - 3 + 4} \Rightarrow \frac{a_{m}}{b_{m}} = \frac{4 m + 1}{6 m + 1}$

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