Question:
Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms
Solution:
Let the common difference of the AP be d.
First term, a = 5
Now,
$a_{1}+a_{2}+a_{3}+a_{4}=\frac{1}{2}\left(a_{5}+a_{6}+a_{7}+a_{8}\right)$ (Given)
$\Rightarrow a+(a+d)+(a+2 d)+(a+3 d)=\frac{1}{2}[(a+4 d)+(a+5 d)+(a+6 d)+(a+7 d)] \quad\left[a_{n}=a+(n-1) d\right]$
$\Rightarrow 4 a+6 d=\frac{1}{2}(4 a+22 d)$
$\Rightarrow 8 a+12 d=4 a+22 d$
$\Rightarrow 22 d-12 d=8 a-4 a$
$\Rightarrow 10 d=4 a$
$\Rightarrow d=\frac{2}{5} a$
$\Rightarrow d=\frac{2}{5} \times 5=2 \quad(a=5)$
Hence, the common difference of the AP is 2.