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8. Sequences and Series
hard
If $\left\{a_{i}\right\}_{i=1}^{n}$ where $n$ is an even integer, is an arithmetic progression with common difference $1$ , and $\sum \limits_{ i =1}^{ n } a _{ i }=192, \sum \limits_{ i =1}^{ n / 2} a _{2 i }=120$, then $n$ is equal to
A
$48$
B
$96$
C
$92$
D
$104$
(JEE MAIN-2022)
Solution
$\sum \limits_{ i =1}^{ n } a _{ i }=\frac{ n }{2}\left\{2 a _{1}+( n +1)\right\}=192$
$\Rightarrow 2 a _{1}+( n -1)=\frac{384}{ n } \ldots-(1)$
$\sum \limits_{ i =1}^{ n / 2} a _{2 i }=\frac{ n }{4}\left[2 a _{1}+2+\left(\frac{ n }{2}-1\right) 2\right]=120$
$2 a _{1}+ n =\frac{480}{ n } \cdots-(2)$
From equation ($2$) and ($1$)
$1=\frac{480}{ n }-\frac{384}{ n }$
$n =480-384=96$
Standard 11
Mathematics
