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:

Step-by-step solution

Given the sum of the first n terms of the A.P. is S_n = 3n^2 + 5n . The n -th term of the A.P. is given by: T_n = S_n - S_ n-1 T_n = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] T_n = 3n^2 + 5n - [3(n^2 - 2n + 1) + 5n - 5] T_n = 3n^2 + 5n - (3n^2 - 6n + 3 + 5n - 5) T_n = 6n + 2 We need to find the sum of the squares of the first 10 terms: _ n=1 ¹⁰ T_n^2 = _ n=1 ¹⁰ (6n + 2)^2 _ n=1 ¹⁰ T_n^2 = _ n=1 ¹⁰ (36n^2 + 24n + 4) _ n=1 ¹⁰ T_n^2 = 36 _ n=1 ¹⁰ n^2 + 24 _ n=1 ¹⁰ n + _ n=1 ¹⁰ 4 Using the formulas for the sum of squares and s