Numerical answer
The sum of the first n terms of an arithmetic progression is given by Sₙ = 3n² + 5n. Find the 15th term of the progression.
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Answer: 92
The nth term is aₙ = Sₙ − Sₙ₋₁ = [3n²+5n] − [3(n−1)²+5(n−1)] = 3n²+5n−3n²+6n−3−5n+5 = 6n+2. So a₁₅ = 6(15)+2 = 92. Shortcut: for Sₙ = An²+Bn, the AP has first term A+B and common difference 2A, so aₙ = (A+B)+(n−1)(2A) = 8+14×6 = 92.
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