Question

The sum of the first n terms of two A.P.'s are as $3n+5:5n-9$. The $4$th terms are equal.

• A. True
• B. False

Answer 1

The correct option is A True

Solution:- (A) True

Let $a$ and $d$ be the first term and common difference of first A.P. and $A$ and $D$ be the first term and common difference of second A.P.

Therefore,

Sum of first $n$ terms of first A.P.-

$S_n=\frac{n}{2}[2a+(n-1)d]$

$a_4=a+3d$

Sum of first $n$ terms of second A.P.-

${S_n}^{\prime }=\frac{n}{2}[2A+(n-1)d]$

$A_4=A+3D$

Given that the ratio of sum of $n$ terms of first and second A.P. is $3n+5:5n-9$.

$\therefore \frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2A+(n-1)D]}=\frac{3n+5}{5n-9}$

$\Rightarrow \frac{2a+(n-1)d}{2A+(n-1)D}=\frac{3n+5}{5n-9}$

$\Rightarrow \frac{a+\left(\frac{n-1}{2}\right)d}{A+\left(\frac{n-1}{2}\right)D}=\frac{3n+5}{5n-9}.....\left(1\right)$

As we need to find the $4^{th}$ term-

$\therefore \frac{n-1}{2}=3$

$\Rightarrow n-1=6$

$\Rightarrow n=7$

Substituting the value of $n$ in ${eq}^n\left(1\right)$, we have

$\frac{a+3d}{A+3D}=\frac{(3\times 7)+5}{(5\times 7)-9}$

$\Rightarrow \frac{a_4}{A_4}=\frac{26}{26}=1$

$\Rightarrow a_4=A_4$

The $4^{th}$ term of the two A.P.'s are equal.

Hence the given statement is true.