Question

The sum of $n$ terms of two $A.P^{\prime }s$ are in the ratio $(5n+21):(3n-13)$. The ratio of $24th$ terms is 1:?

Answer 1

Let the two A.P's be $a,a+d,a+2d,....$ and $a^{\prime },a^{\prime }+d^{\prime },a^{\prime }+2d^{\prime },....$
Let their sum be $S_n$ and $S_n^{\prime }$ respectively.
$\therefore S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ and $S_n^{\prime }=\frac{n}{2}[2a^{\prime }+(n-1\left)d^{\prime }\right]$
$\because \frac{S_n}{S_n^{\prime }}=\frac{3n-13}{5n+21}$
$\therefore \frac{a+\left(\frac{n-1}{2}\right)d}{a^{\prime }+\left(\frac{n-1}{2}\right)d^{\prime }}=\frac{3n-13}{5n+21}\cdots \left(1\right)$
The $24$th terms of two A.P's are respectively
$T_{24}=a+23d$ and $T_{24}=a^{\prime }+23d^{\prime }$
$\therefore \frac{T_{24}}{T_{24}^{\prime }}=\frac{a+23d}{a^{\prime }+23d^{\prime }}\cdots \left(2\right)$
Comparing $\left(1\right)$ and $\left(2\right)$, we get
$\frac{n-1}{2}=23$
$\therefore n=47$
from $\left(1\right),\frac{a+23d}{a^{\prime }+23d^{\prime }}=\frac{3\times 47-13}{5\times 47+21}$
$=\frac{1}{2}$
$\therefore \frac{T_{24}}{T_{24}^{\prime }}=\frac{1}{2}$
$\Rightarrow T_{24}:T_{24}^{\prime }=1:2$