Let S_n be the sum to n-terms of an arithmetic progression 3,7,11, If 40

Question

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

Answer: 9

Accepted Answer

$\begin{gathered} {S}_{n}=3+7+11+\ldots \ldots n \ =\frac{n}{2}(6+(n-1)4)=3n+2{n}^{2}-2n \ =2{n}^{2}+n \ \sum _{k=1}^{n}{S}_{k}=2\sum _{k=1}^{n}{K}^{2}+\sum _{k=1}^{n}K \ =2\cdot \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \ =n(n+1)[\frac{2n+1}{3}+\frac{1}{2}] \ =\frac{n(n+1)(4n+5)}{6} \ \Rightarrow 40<\frac{6}{n(n+1)}\sum _{k=1}^{n}{S}_{k}<42 \ 40<4n+5<42 \ 35<4n<37 \ n=9 \end{gathered}$

Solutions · 1