Numerical answer
The sum of the first n terms of an arithmetic progression is 3n² + 5n for every natural number n. What is the 20th term of the progression?
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Answer: 122
The nth term Tₙ = Sₙ − Sₙ₋₁.
S₂₀ = 3(400) + 5(20) = 1300; S₁₉ = 3(361) + 5(19) = 1083 + 95 = 1178.
T₂₀ = 1300 − 1178 = 122.
General form: Tₙ = 3n² + 5n − [3(n − 1)² + 5(n − 1)] = 6n + 2 → T₂₀ = 122. (First term 8, common difference 6.)
S₂₀ = 3(400) + 5(20) = 1300; S₁₉ = 3(361) + 5(19) = 1083 + 95 = 1178.
T₂₀ = 1300 − 1178 = 122.
General form: Tₙ = 3n² + 5n − [3(n − 1)² + 5(n − 1)] = 6n + 2 → T₂₀ = 122. (First term 8, common difference 6.)
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