Question

Let the sum of the first n terms of an A.P. be $3n^2 + 5n$. Then the sum of squares of the first 10 terms of the A.P. is:

• A. 10220
• B. 12860
• C. 15220
• D. 19780

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the formula for the sum of the first $n$ terms of a sequence, $S_n = 3n^2 + 5n$. We must first find the general term of this sequence ($a_n$) and then compute the sum of the squares of its first 10 terms, i.e., $\sum_{k=1}^{10} a_k^2$.
Step 2: Key Formula or Approach:
1. The general term $a_n$ can be found from the sum formula $S_n$ by using the relation $a_n = S_n - S_{n-1}$ (for $n \ge 2$) and $a_1 = S_1$.
2. Once $a_n$ is determined, we will set up the sum $\sum_{k=1}^{10} a_k^2$.
3. This sum will be evaluated using standard summation formulas:
- $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
- $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
- $\sum_{k=1}^{n} (\text{constant}) = n \times (\text{constant})$
Step 3: Detailed Explanation:
The sum of the first $n$ terms is given as $S_n = 3n^2 + 5n$.
First, we find the general term $a_n$.
The first term is $a_1 = S_1 = 3(1)^2 + 5(1) = 8$.
For $n \ge 2$, we have:
\[ a_n = S_n - S_{n-1} \] \[ a_n = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] \] \[ a_n = (3n^2 + 5n) - [3(n^2 - 2n + 1) + 5n - 5] \] \[ a_n = (3n^2 + 5n) - [3n^2 - 6n + 3 + 5n - 5] \] \[ a_n = (3n^2 + 5n) - (3n^2 - n - 2) \] \[ a_n = 6n + 2 \] Let's check if this formula holds for $n=1$: $a_1 = 6(1) + 2 = 8$. This matches $S_1$, so the formula is valid for all $n \ge 1$.
Now, we compute the sum of the squares of the first 10 terms:
\[ \sum_{k=1}^{10} a_k^2 = \sum_{k=1}^{10} (6k + 2)^2 \] First, expand the expression inside the sum:
\[ (6k+2)^2 = 36k^2 + 24k + 4 \] The summation becomes:
\[ \sum_{k=1}^{10} (36k^2 + 24k + 4) = 36\sum_{k=1}^{10} k^2 + 24\sum_{k=1}^{10} k + \sum_{k=1}^{10} 4 \] We apply the summation formulas with $n=10$:
\[ \sum_{k=1}^{10} k = \frac{10(11)}{2} = 55 \] \[ \sum_{k=1}^{10} k^2 = \frac{10(11)(21)}{6} = 385 \] \[ \sum_{k=1}^{10} 4 = 10 \times 4 = 40 \] Substituting these results back:
\[ \text{Sum} = 36(385) + 24(55) + 40 \] \[ \text{Sum} = 13860 + 1320 + 40 = 15220 \] Step 4: Final Answer:
The sum of squares of the first 10 terms of the A.P. is 15220.