Jee Main 2024 30 01 2024 Shift 2 - Question9
Question 9
If $S _n=3+7+11+\ldots$ upto $n$ terms
And $40<\frac{6}{n(n+1)} \sum _{k=1}^{n} S _k<45$, then $n$ is
(1) 9
(2) 10
(3) 11
(4) 12
Show Answer
Answer: (1)
Solution:
$S _n=n(2 n+1)$
$$ \begin{aligned} & \sum _{k=1}^{n} S _k=\sum _{k=1}^{n}\left(2 k^{2}+k\right) \\ & \quad=2 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} \\ & \therefore \frac{6}{n(n+1)} \sum _{k=1}^{n} S _k \\ & =\frac{6}{n(n+1)} \cdot n(n+1)\left(\frac{2 n+1}{3}+\frac{1}{2}\right) \\ & =4 n+2+3 \\ & =4 n+5 \\ & \because 40<\frac{6}{n(n+1)} \sum _{k=1}^{n} S _k<45 \\ & 40<4 n+5<45 \\ & 3 s<4 n<40 \\ & \therefore n=9 \end{aligned} $$
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