Sequences And Series Question 15
Question 15 - 2024 (30 Jan Shift 2)
Let $S_{n}$ be the sum to n-terms of an arithmetic progression $3,7,11, \ldots \ldots$
If $40<\left(\frac{6}{n(n+1)} \sum_{k=1}^{n} s_{k}\right)<42$, then $n$ equals
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Answer (9)
Solution
$S_{n}=3+7+11+\ldots \ldots . n$ terms
$$ \begin{aligned} & =\frac{n}{2}(6+(n-1) 4)=3 n+2 n^{2}-2 n \ & =2 n^{2}+n \ & \sum_{k=1}^{n} S_{k}=2 \sum_{k=1}^{n} K^{2}+\sum_{k=1}^{n} K \ & =2 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} \ & =n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right] \ & =\frac{n(n+1)(4 n+5)}{6} \ & \Rightarrow 40<\frac{6}{n(n+1)} \sum_{k=1}^{n} S_{k}<42 \ & 40<4 n+5<42 \ & 35<4 n<37 \ & n=9 \end{aligned} $$
