Numerical answer
The sum of the first n terms of an arithmetic progression is 3n² + 5n for every positive integer n. What is the 15th term of the progression?
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Answer: 92
The nth term is Tₙ = Sₙ − Sₙ₋₁.
T₁₅ = S₁₅ − S₁₄ = (3·225 + 75) − (3·196 + 70) = 750 − 658 = 92.
In general Tₙ = 6n + 2, so the AP is 8, 14, 20, … with common difference 6; T₁₅ = 6·15 + 2 = 92.
Tip: If Sₙ = pn² + qn, then the common difference is 2p and the first term is p + q.
T₁₅ = S₁₅ − S₁₄ = (3·225 + 75) − (3·196 + 70) = 750 − 658 = 92.
In general Tₙ = 6n + 2, so the AP is 8, 14, 20, … with common difference 6; T₁₅ = 6·15 + 2 = 92.
Tip: If Sₙ = pn² + qn, then the common difference is 2p and the first term is p + q.
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