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. [CBSE 2012]

Answer 1

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)
$\Rightarrow a+\left(a+d\right)+\left(a+2d\right)+\left(a+3d\right)=\frac{1}{2}\left[\left(a+4d\right)+\left(a+5d\right)+\left(a+6d\right)+\left(a+7d\right)\right]$ $\left[a_n=a+\left(n-1\right)d\right]$
$$\begin{gathered} \Rightarrow 4a+6d=\frac{1}{2}\left(4a+22d\right) \\ \Rightarrow 8a+12d=4a+22d \\ \Rightarrow 22d-12d=8a-4a \\ \Rightarrow 10d=4a \end{gathered}$$
$$\begin{gathered} \Rightarrow d=\frac{2}{5}a \\ \Rightarrow d=\frac{2}{5}\times 5=2\left(a=5\right) \end{gathered}$$

Hence, the common difference of the AP is 2.