Question

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

Answer 1

Identify the arithmetic progression: First term $a = 3$, common difference $d = 4$.

Find the sum of the first $n$ terms $S_n$:

$S_n = \frac{n}{2} \left( 2a + (n - 1)d \right) = \frac{n}{2} \left(4n + 2\right) = n(2n + 1).$

Calculate $\sum_{k=1}^{n} S_k$:

$\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} k(2k + 1) = 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k.$

Using formulas:

$\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}, \quad \sum_{k=1}^{n} k = \frac{n(n + 1)}{2},$

thus,

$\sum_{k=1}^{n} S_k = \frac{n(n + 1)(4n + 5)}{6}.$

Set up the inequality:

\frac{6}{n(n + 1)} \times \frac{n(n + 1)(4n + 5)}{6} < 42\.

Simplifying gives:

$4n + 5 < 42 \implies 4n < 37 \implies n < 9.25.$

Find the largest integer $n$:

$n = 9.$
Thus, $n = 9$.