8. Sequences and Series
medium

અહી $S_{n}$ એ સમાંતર શ્રેણીના પ્રથમ $n$ પદોનો સરવાળો દર્શાવે છે. જો $S_{3 n}=3 S_{2 n}$ હોય તો $\frac{S_{4 n}}{S_{2 n}}$ ની કિમંત મેળવો.

A

$4$

B

$6$

C

$8$

D

$2$

(JEE MAIN-2021)

Solution

Let a be first term and $d$ be common diff. of this A.P.

Given $\mathrm{S}_{3 \mathrm{n}}=3 \mathrm{~S}_{2 \mathrm{n}}$

$\Rightarrow \frac{3 n}{2}[2 a+(3 n-1) d]=3 \frac{2 n}{2}[2 a+(2 n-1) d]$

$\Rightarrow 2 a+(3 n-1) d=4 a+(4 n-2) d$

$\Rightarrow 2 a+(n-1) d=0$

$\text { Now } \frac{S_{a n}}{S_{2 n}}=\frac{\frac{4 n}{2}[2 a+(4 n-1) d]}{\frac{2 n}{2}[2 a+(2 n-1) d]}=\frac{2[\underbrace{2 a+(n-1) d}_{-0}+3 n d]}{[\underbrace{2 a+(n-1) d}_{-0}+n d]}$

$=\frac{6 n d}{n d}=6$

Standard 11
Mathematics

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