Question

If ${\left\{a_i\right\}}_{i=1}^n$ where $n$ is an even integer , is an arithmetic progression with common difference $1$ , and $\sum _{i=1}^n{a}_i=192,\sum _{i=1}^{n/2}{a}_{2i}=120$, then $n$ is equal to:

• A. 48
• B. 96
• C. 92
• D. 104

Answer 1

Answer: (B)

$\sum _{i=1}^n{a}_i=\frac{n}{2}\left\{2a_1+(n+1)\right\}=192$
$\Rightarrow 2a_1+(n-1)=\frac{384}{n}$ .......(1)
$\sum _{i=1}^{n/2}{a}_{2i}=\frac{n}{4}\left[2a_1+2+\left(\frac{n}{2}-1\right)2\right]=120$
$2a_1+n=\frac{480}{n}$ ......(2)
From equation (2) and (1)
$1=\frac{480}{n}-\frac{384}{n}$
$n=480-384=96$