Let the sum of the first n terms of an A.P. b | vedclass

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(C) Given the sum of the first $n$ terms $S_n = 3n^2 + 5n$.The $n^{th}$ term $T_n$ is given by $T_n = S_n - S_{n-1}$.$T_n = (3n^2 + 5n) - (3(n-1)^2 +

Vedclass · Accepted Answer

$15220$

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(C) Given the sum of the first $n$ terms $S_n = 3n^2 + 5n$.The $n^{th}$ term $T_n$ is given by $T_n = S_n - S_{n-1}$.$T_n = (3n^2 + 5n) - (3(n-1)^2 + 5(n-1)) = 3n^2 + 5n - (3(n^2 - 2n + 1) + 5n - 5) = 3n^2 + 5n - (3n^2 - 6n + 3 + 5n - 5) = 6n + 2$.The first $10$ terms are $8, 14, 20, \dots, 62$.We need to find the sum of squares: $\sum_{n=1}^{10} (6n + 2)^2$.$= \sum_{n=1}^{10} (36n^2 + 24n + 4) = 36 \sum_{n=1}^{10} n^2 + 24 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 4$.Using standard summation formulas: $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$.$= 36 \left( \frac{10 \cdot 11 \cdot 21}{6} \right) + 24 \left( \frac{10 \cdot 11}{2} \right) + 4(10)$.$= 6(10 \cdot 11 \cdot 21) + 12(110) + 40$.$= 6(2310) + 1320 + 40 = 13860 + 1360 = 15220$.

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:

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