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

To find the sum of the squares of the first 10 terms, we first need to identify the individual terms of the Arithmetic Progression (A.P.).

The sum of the first $$n$$ terms is given by $$S_n = 3n^2 + 5n$$. We can find the $$n^{th}$$ term using the formula $$a_n = S_n - S_{n-1}$$:

$$a_n = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)]$$

$$a_n = (3n^2 + 5n) - [3(n^2 - 2n + 1) + 5n - 5]$$

$$a_n = 3n^2 + 5n - [3n^2 - 6n + 3 + 5n - 5]$$

$$a_n = 3n^2 + 5n - 3n^2 + n + 2$$

$$a_n = 6n + 2$$

Substituting $$n=1, 2, 3...$$ into $$a_n = 6n + 2$$:

• $$a_1 = 6(1) + 2 = 8$$
• $$a_2 = 6(2) + 2 = 14$$
• $$a_3 = 6(3) + 2 = 20$$
• $$\sum n^2 = \frac{n(n+1)(2n+1)}{6}$$
• $$\sum n = \frac{n(n+1)}{2}$$
• $$\sum_{n=1}^{10} n^2 = \frac{10 \times 11 \times 21}{6} = 385$$
• $$\sum_{n=1}^{10} n = \frac{10 \times 11}{2} = 55$$
• $$\sum_{n=1}^{10} 4 = 4 \times 10 = 40$$

The common difference ($$d$$) is 6.

We need to find $$\sum_{n=1}^{10} (a_n)^2 = \sum_{n=1}^{10} (6n + 2)^2$$.

First, expand the square:

$$(6n + 2)^2 = 36n^2 + 24n + 4$$

Now, apply the summation from $$n=1$$ to $$10$$:

$$\text{Sum} = 36 \sum_{n=1}^{10} n^2 + 24 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 4$$

Use the standard summation formulas:

For $$n=10$$:

$$\text{Sum} = 36(385) + 24(55) + 40$$

$$\text{Sum} = 13860 + 1320 + 40$$

$$\text{Sum} = 15220$$

Final Answer: 15220 (Option C)