If the sums of n terms of two APs are in ratio

Solution:

Given: sums of n terms of two APs are in ratio (2n + 3) : (3n + 2)

To find: find the ratio of their 10th terms

For the sum of n terms of two APs is given by

$\mathrm{S}_{1}=\frac{\mathrm{n}}{2}\left(2 \mathrm{a}_{1}+(\mathrm{n}-1) \times \mathrm{d}_{1}\right)$

$\mathrm{S}_{2}=\frac{\mathrm{n}}{2}\left(2 \mathrm{a}_{2}+(\mathrm{n}-1) \times \mathrm{d}_{2}\right)$

$\frac{\mathrm{S}_{1}}{\mathrm{~S}_{2}}=\frac{2 \mathrm{n}+3}{3 \mathrm{n}+2}$

$=\frac{\left(2 \mathrm{a}_{1}+(\mathrm{n}-1) \times \mathrm{d}_{1}\right.}{\left(2 \mathrm{a}_{2}+(\mathrm{n}-1) \times \mathrm{d}_{2}\right)}$

Or we can write it as

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

For $10^{\text {th }}$ term put $\frac{(\mathrm{n}-1)}{2}=10$

n = 19

Therefore the ratio of the 10th term will be

$=\frac{2 \times 19+3}{3 \times 19+2}$

$\Rightarrow 41: 57$