The sum of the first n terms of an arithmetic progression is given by Sₙ = 3n² + 5n.

The nth term of a series equals Sₙ − Sₙ₋₁. Here Sₙ = 3n²+5n, so Tₙ = [3n²+5n] − [3(n−1)²+5(n−1)] = 3n²+5n − 3n²+6n−3−5n+5 = 6n+2. So T₁₀ = 6(10)+2 = 62. Shortcut: for Sₙ = An²+Bn, the AP has first term A+B and common difference 2A, so Tₙ = (A+B) + (n−1)(2A) = 6n+2 here too — same result.