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

Correct Answer: 9

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, \( n = 9 \).