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 :
Correct Option: 1
Solution:
Given $a=3$ and $S_{25}=S_{40}-S_{25}$
$\Rightarrow 2 S_{25}=S_{40}$
$\Rightarrow 2 \times \frac{25}{2}[6+24 d]=\frac{40}{2}[6+39 d]$
$\Rightarrow 25[6+24 d]=20[6+39 d]$
$\Rightarrow 5(2+8 d)=4(2+13 d)$
$\Rightarrow 10+40 d=8+52 d$
$\Rightarrow d=\frac{1}{6}$