Sum of Squares of First 10 Terms of Arithmetic Progression

To find the sum of the squares of the first 10 terms of the given Arithmetic Progression (A.P.), we first need to determine the general term, $a_n$, of the sequence.

Given that the sum of the first $n$ terms of the A.P. is: $S_n = 3n^2 + 5n$

The $n$-th term $a_n$ can be found using the relation $a_n = S_n - S_{n-1}$ for $n \ge 1$: $a_n = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)]$

Expanding the expression: $a_n = 3n^2 + 5n - [3(n^2 - 2n + 1) + 5n - 5]$ $a_n = 3n^2 + 5n - (3n^2 - 6n + 3 + 5n - 5)$ $a_n = 3n^2 + 5n - (3n^2 - n - 2)$ $a_n = 6n + 2$

Now, we need to calculate the sum of the squares of the first 10 terms, given by: $S = \sum_{n=1}^{10} a_n^2 = \sum_{n=1}^{10} (6n + 2)^2$

Expanding the term inside the summation: $(6n + 2)^2 = 36n^2 + 24n + 4$

Thus, the sum can be split into individual summations: $S = 36 \sum_{n=1}^{10} n^2 + 24 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 4$

Using the standard summation formulas for $N = 10$:

1. $\sum_{n=1}^{N} n = \frac{N(N+1)}{2} = \frac{10 \times 11}{2} = 55$
2. $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6} = \frac{10 \times 11 \times 21}{6} = 385$
3. $\sum_{n=1}^{10} 4 = 4 \times 10 = 40$

Substituting these values back into the expression for $S$: $S = 36(385) + 24(55) + 40$ $S = 13860 + 1320 + 40$ $S = 15220$

Thus, the sum of the squares of the first 10 terms of the A.P. is 15220.