Question

The sums of n terms of two A.P.'s are in the ratio $3n+2:2n+3$. Find the ratio of their ${10}^{th}$ terms.

Answer 1

Let $a,a+d,a+2d,a+3d,.......$
$A,A+D,A+2D,A+3D,........$
be two A.P.'s $\frac{\frac{n}{2}[2a+(n-1\left)d\right]}{\frac{n}{2}[2A+(n-1\left)D\right]}=\frac{3n+2}{2n+3}$ (given)
$\Rightarrow \frac{a+\frac{n-1}{2}d}{A+\frac{n-1}{2}D}=\frac{3n+2}{2n+3}$
$\Rightarrow$ To get the ratio ${10}^{th}$ terms put $\frac{n-1}{2}=9$ or $n=19$
$\Rightarrow \frac{a+9d}{A+9D}=\frac{3\left(19\right)+2}{2\left(19\right)+3}=\frac{59}{41}$