Sequences & Series
AP — Finding n from Inequality on Partial Sum
nta_pyq_2024_jan
Grade 11
Question:
Let $S_n$ be the sum to $n$-terms of an arithmetic progression $3,7,11,\ldots$ If $40<\dfrac{6}{n(n+1)}\displaystyle\sum_{k=1}^{n}S_k<42$, then $n$ equals
Step-by-Step Solution
Key Concept: $S_k=2k^2+k$. $\sum_{k=1}^n S_k=2\sum k^2+\sum k=\frac{n(n+1)(4n+5)}{6}$. Substituting: $40<4n+5<42\Rightarrow35<4n<37$.
$4n+5=4(9)+5=41\in(40,42)$. $n=9$.
Correct Answer: 9