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

Given an arithmetic progression with first term \( a = 3 \) and common difference \( d = 4 \).

The sum of the first \( n \) terms is \( S_n \):

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

We need to 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 the standard summation formulas:

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

The sum is:

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

We are given the inequality:

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

Simplifying the inequality yields:

\[ 4n + 5 < 42 \implies 4n < 37 \implies n < 9.25. \]

The largest integer \( n \) satisfying this condition is:

\[ n = 9. \]
Therefore, \( n = 9 \).