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, \( n = 9 \).

Answer 2

The problem asks to find the integer value of \( n \) that satisfies a given inequality. The inequality involves the sum of the sums of the first \( k \) terms of a given arithmetic progression (AP), for \( k \) from 1 to \( n \).

Concept Used:

1. Arithmetic Progression (AP): The sum of the first \( n \) terms of an AP with first term \( a \) and common difference \( d \) is given by the formula:

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

2. Summation Formulas: We will use the standard formulas for the sum of the first \( n \) natural numbers and the sum of the squares of the first \( n \) natural numbers:

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

Step-by-Step Solution:

Step 1: Identify the parameters of the given arithmetic progression.

The AP is \( 3, 7, 11, \ldots \).

The first term is \( a = 3 \).

The common difference is \( d = 7 - 3 = 4 \).

Step 2: Find a general formula for \( S_k \), the sum of the first \( k \) terms of this AP.

\[ S_k = \frac{k}{2}[2a + (k-1)d] \]

Substitute the values of \( a \) and \( d \):

\[ S_k = \frac{k}{2}[2(3) + (k-1)4] = \frac{k}{2}[6 + 4k - 4] = \frac{k}{2}(4k + 2) \] \[ S_k = k(2k + 1) = 2k^2 + k \]

Step 3: Calculate the sum \( \sum_{k=1}^{n} S_k \).

\[ \sum_{k=1}^{n} S_k = \sum_{k=1}^{n} (2k^2 + k) = 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \]

Now, substitute the standard summation formulas:

\[ \sum_{k=1}^{n} S_k = 2 \left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2} \]

Simplify the expression by taking \( \frac{n(n+1)}{2} \) as a common factor:

\[ \sum_{k=1}^{n} S_k = \frac{n(n+1)}{2} \left[ 2 \frac{(2n+1)}{3} + 1 \right] \] \[ = \frac{n(n+1)}{2} \left[ \frac{4n+2+3}{3} \right] = \frac{n(n+1)(4n+5)}{6} \]

Step 4: Substitute the result of the summation into the expression from the inequality.

The expression is \( E = \frac{6}{n(n+1)} \sum_{k=1}^{n} S_k \).

\[ E = \frac{6}{n(n+1)} \left( \frac{n(n+1)(4n+5)}{6} \right) \] \[ E = 4n + 5 \]

Step 5: Solve the given inequality for \( n \).

The inequality is \( 40 < E < 42 \).

\[ 40 < 4n + 5 < 42 \]

Subtract 5 from all parts of the inequality:

\[ 40 - 5 < 4n < 42 - 5 \] \[ 35 < 4n < 37 \]

Divide all parts by 4:

\[ \frac{35}{4} < n < \frac{37}{4} \] \[ 8.75 < n < 9.25 \]

Final Computation & Result:

Since \( n \) represents the number of terms, it must be an integer. The only integer value of \( n \) that lies in the interval \( (8.75, 9.25) \) is 9.

Therefore, the value of \( n \) is 9.