If the sums of n terms of two arithmetic progressions are in the ratio 2n + 5 :

Given:

$\frac{S_{n}}{S_{n}{ }^{1}}=\frac{2 n+5}{3 n+4}$

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

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

Ratio of their $\mathrm{m}$ terms $=\frac{a_{m}}{b_{m}}$

To find the ratio of the mth terms, replace n by 2m-">−-1 in equation (1).

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

$\Rightarrow \frac{a+(m-1) d}{b+(m-1) d^{1}}=\frac{4 m-2+3}{6 m-3+4}$

$\Rightarrow \frac{a_{m}}{b_{m}}=\frac{4 m+1}{6 m+1}$