If the sums of n terms of two arithmetic progressions are the ratio (2n + 3) :

Solution:

Let a1 and d2 represent the first term and common difference of first arithmetic progression.

Also, a2 and d2 represent the first term  and common difference of second arithmetic progression then sum of n terms of first A.P

$\left(S_{n}\right)_{1}=\frac{n}{2}\left\{2 a_{1}+(n-1) d_{1}\right\}$

and sum of n terms of second A.P

$\left(S_{n}\right)_{2}=\frac{n}{2}\left\{2 a_{2}+(n-1) d_{2}\right\}$

Now, according to given condition

$\frac{\left(S_{n}\right)_{1}}{\left(S_{n}\right)_{0}}=\frac{2 n+3}{6 n+5}$

i. e $\frac{2 n+3}{6 n+5}=\frac{\frac{n}{2}\left\{2 a_{1}+(n-1) d_{1}\right\}}{\frac{n}{2}\left\{2 a_{2}+(n-1) d_{2}\right\}}$

$\frac{2 n+3}{6 n+5}=\frac{a_{1}+\frac{(n-1)}{2} d_{1}}{a_{2}+\frac{(n-1)}{2} d_{2}}$

Since $n^{\text {th }}$ term of any A.P is $a+(n-1) d$

$\therefore 13^{\text {th }}$ term of first A.P is $a_{1}+12 d_{1}$ and $13^{\text {th }}$ term of second A.P is $a_{2}+12 d_{2}$

i. e $\frac{n-1}{2}=12$

i. e $\frac{n-1}{2}=12$

i. e $n=24+1$

i. e $n=25$

$\therefore \frac{a_{1}+12 d_{1}}{a_{2}+12 d_{2}}=\left(\frac{2 n+3}{6 n+5}\right)$ at $n=25$

$=\frac{2(25)+3}{6(25)+5}$

$\therefore \frac{\left(a_{13}\right)_{1}}{\left(a_{13}\right)_{2}}=\frac{53}{155}$

i.e ratio of their $13^{\text {th }}$ terms is $\frac{53}{155}$.