Let Sₙ be sum to n-terms of an arithmetic progression 3,7,11, ldots ldots . . If 40<left(frac{6

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8. Sequences and Series

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Let $S_n$ be the sum to n-terms of an arithmetic progression $3,7,11, \ldots \ldots$. . If $40<\left(\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}\right)<42$, then $\mathrm{n}$ equals

A

$9$

B

$8$

C

$10$

D

$7$

(JEE MAIN-2024)

Solution

$S_n= 3+7+11+………… n $ terams

${n}{2}(6+(n-1) 4)=3 n+2 n^2-2 n $

$ =2 n^2+n $

$ \sum_{k=1}^n S_k=2 \sum_{k=1}^n K^2+\sum_{k=1}^n K $

$ =2 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} $

$=n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right]$

$=\frac{n(n+1)(4 n+5)}{6} $

Rightarrow $40<\frac{6}{n(n+1)} \sum_{k=1}^n S_k<42 $

$ 40<4 n+5<42 $

$ 35<4 n<37 $

$ n=9$

Standard 11

Mathematics

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