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

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