Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence of numbers such that the difference between the consecutive terms is constant. The sum of the first $n$ terms of an AP is given by $S_n = \frac{n}{2}[2a + (n-1)d]$, where $a$ is the first term and $d$ is the common difference.
• Summation Formulas:
  • Sum of first $n$ natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
  • Sum of first $n$ squares: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
• Inequality Solving: Manipulating inequalities by performing the same operation on all parts.

Step-by-Step Solution

Step 1: Determine the sum of $n$ terms of the given AP, $S_n$. The given arithmetic progression is $3, 7, 11, \ldots$. The first term is $a = 3$. The common difference is $d = 7 - 3 = 4$. We use the formula for the sum of $n$ terms of an AP: $S_n = \frac{n}{2}[2a + (n-1)d]$ Substituting the values of $a$ and $d$: $S_n = \frac{n}{2}[2(3) + (n-1)4]$ $S_n = \frac{n}{2}[6 + 4n - 4]$ $S_n = \frac{n}{2}[4n + 2]$ Factoring out 2 from the bracket: $S_n = \frac{n}{2} \cdot 2(2n + 1)$ $S_n = n(2n + 1) = 2n^2 + n$ This expression gives the sum of the first $n$ terms of the AP.

Step 2: Calculate the sum $\sum_{k=1}^{n} S_k$. We need to find the sum of $S_k = 2k^2 + k$ from $k=1$ to $n$. $\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} (2k^2 + k)$ Using the linearity of summation: $\sum_{k=1}^{n} S_k = 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$ Applying the standard summation formulas: $\sum_{k=1}^{n} S_k = 2 \left( \frac{n(n+1)(2n+1)}{6} \right) + \left( \frac{n(n+1)}{2} \right)$ Simplify the expression: $\sum_{k=1}^{n} S_k = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$ Find a common denominator (6) and factor out $n(n+1)$: $\sum_{k=1}^{n} S_k = n(n+1) \left[ \frac{2n+1}{3} + \frac{1}{2} \right]$ Combine the fractions inside the bracket: $\sum_{k=1}^{n} S_k = n(n+1) \left[ \frac{2(2n+1) + 3(1)}{6} \right]$ $\sum_{k=1}^{n} S_k = n(n+1) \left[ \frac{4n+2 + 3}{6} \right]$ $\sum_{k=1}^{n} S_k = \frac{n(n+1)(4n+5)}{6}$

Step 3: Solve the given inequality for $n$. The given inequality is: $40 < \left(\frac{6}{n(n+1)} \sum_{k=1}^n S_k\right) < 42$ Substitute the expression for $\sum_{k=1}^n S_k$: $40 < \left(\frac{6}{n(n+1)} \cdot \frac{n(n+1)(4n+5)}{6}\right) < 42$ Cancel out the common terms: $40 < (4n+5) < 42$ Subtract 5 from all parts of the inequality: $40 - 5 < 4n < 42 - 5$ $35 < 4n < 37$ Divide all parts by 4: $\frac{35}{4} < n < \frac{37}{4}$ Convert to decimals: $8.75 < n < 9.25$ Since $n$ must be a natural number (positive integer), the only integer value that satisfies this inequality is $n=9$.

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when simplifying fractions and combining terms.
• Formula Application: Ensure correct recall and application of the summation formulas for $k$ and $k^2$.
• Integer Constraint: Remember that $n$ represents the number of terms, so it must be a positive integer. This is crucial for selecting the correct value from an inequality.

Summary

The problem requires us to first find the general term for the sum of an arithmetic progression ($S_n$), then compute the sum of these terms up to $n$ ($\sum_{k=1}^{n} S_k$), and finally use the given inequality to solve for $n$. By carefully applying the formulas for AP sums and summation of powers, and performing accurate algebraic simplification, we arrived at the inequality $8.75 < n < 9.25$. Given that $n$ must be a natural number, the unique solution is $n=9$.

The final answer is $\boxed{9}$.