Question

Let \( S_n \) be the sum of the first \( n \) terms of an A.P. If \( S_n = 3n^2 + 5n \), then the sum of the squares of the first ten terms of the given A.P. is:

• A. 15110
• B. 15220
• C. 14202
• D. 14308

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the sum of the squares of the terms, we first need to identify the general term (\( a_n \)) of the Arithmetic Progression. The \( n^{th} \) term can be found using the relation \( a_n = S_n - S_{n-1} \). Once the terms are identified, we apply the summation formulas for \( n^2 \) and \( n \).
Step 2: Key Formula or Approach:
1. \( a_n = S_n - S_{n-1} \).
2. Sum of first \( n \) natural numbers: \( \sum n = \frac{n(n+1)}{2} \).
3. Sum of squares of first \( n \) natural numbers: \( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \).
Step 3: Detailed Explanation:
1. Find the general term \( a_n \): \[ S_n = 3n^2 + 5n \] \[ S_{n-1} = 3(n-1)^2 + 5(n-1) = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - n - 2 \] \[ a_n = (3n^2 + 5n) - (3n^2 - n - 2) = 6n + 2 \] 2. We need to find \( \sum_{n=1}^{10} a_n^2 \): \[ \sum_{n=1}^{10} (6n + 2)^2 = \sum_{n=1}^{10} (36n^2 + 24n + 4) \] 3. Apply summation: \[ = 36 \sum_{n=1}^{10} n^2 + 24 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 4 \] \[ = 36 \left( \frac{10 \times 11 \times 21}{6} \right) + 24 \left( \frac{10 \times 11}{2} \right) + 40 \] \[ = 36(385) + 24(55) + 40 \] \[ = 13860 + 1320 + 40 = 15220 \] (Note: Re-calculating \( a_1 = 8, a_2 = 14 \dots \). If the sum is for 10 terms, the calculation \( 13860 + 1320 + 40 = 15220 \) is correct. Let's re-verify the question's arithmetic options. If \( a_n = 6n+2 \), the sum is 15220. If \( a_n = 6n-2 \), the result would be 15110). Re-check: \( a_n = S_n - S_{n-1} \). For \( n=1, a_1 = S_1 = 8 \). For \( n=1, 6(1)+2 = 8 \). The general term is correct.
Step 4: Final Answer:
The sum of the squares of the first ten terms is 15220.