Question

If the first term of an A.P. is $3$ and the sum of its first $25$ terms is equal to the sum of its next $15$ terms, then the common difference of this A.P. is

• A. $\frac{1}{6}$
• B. $\frac{1}{5}$
• C. $\frac{1}{4}$
• D. $\frac{1}{7}$

Answer 1

The correct option is A

$\frac{1}{6}$

Explanation for the correct option:

Step 1: Solve for sum of first $25$ terms

Given that, first term of an A.P. is $a=3$

Let $n$ be the number of terms and $d$ be the common difference

We know that sum of $n$ terms$=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

$\therefore$Sum of first $25$ terms$\begin{array}{cc}=& \frac{25}{2}\left(2\left(3\right)+\left(25-1\right)d\right)\end{array}$

$\begin{array}{cc}=& \frac{25}{2}\left(6+24d\right)\end{array}$

Step 2: Solve for sum of next $15$ terms

Here the first term will be ${26}^{th}$term i.e. $a+(26-1)d=a+25d$

$\therefore$Sum of next $15$ terms$\begin{array}{cc}=& \frac{15}{2}\left(2\left(a+25d\right)+\left(15-1\right)d\right);\left[\because S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\right]\end{array}$

$\begin{array}{cl}=& \frac{15}{2}\left(2\left(3\right)+50d+14d\right)\\ =& \frac{15}{2}\left(6+64d\right)\end{array}$

Step 3: Solve for common difference

Given that, sum of its first $25$ terms is equal to the sum of its next $15$ terms

$\begin{array}{crcl}\Rightarrow & \frac{25}{2}\left(6+24d\right)& =& \frac{15}{2}\left(6+64d\right)\\ \Rightarrow & 5\left(6+24d\right)& =& 3\left(6+64d\right)\\ \Rightarrow & 30+120d& =& 18+192d\\ \Rightarrow & \left(192-120\right)d& =& 30-18\\ \Rightarrow & 72d& =& 12\\ \Rightarrow & d& =& \frac{12}{72}\\ \Rightarrow & d& =& \frac{1}{6}\end{array}$

Hence, the correct answer is option (A) i.e. $\frac{1}{6}$.