Question

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

• A. 48
• B. 96
• C. 92
• D. 104

Answer 1

Answer: (B)

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