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8. Sequences and Series
hard
यदि $\left\{ a _{ i }\right\}_{ i =1}^{ n }$ (जहाँ $n$ सम पूर्णांक है) समान्तर श्रेढ़ी है जिसका सार्वअन्तर $1$ तथा $\sum \limits_{ i =1}^{ n } a _{ i }=192$, $\sum \limits_{ i =1}^{ n / 2} a _{2 i }=120$ है, तो $n$ बराबर है:
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