Single correct option
The sum of the first n terms of an arithmetic progression is 3n² + 5n for every n. What is its 15th term?
- A. 86
- B. 90
- C. 92
- D. 98
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Answer: (C) 92
The nth term is Tₙ = Sₙ − Sₙ₋₁.
Sₙ − Sₙ₋₁ = 3n² + 5n − [3(n − 1)² + 5(n − 1)] = 3(2n − 1) + 5 = 6n + 2.
T₁₅ = 6 × 15 + 2 = 92.
Check: T₁ = S₁ = 8 = 6(1) + 2 ✓; the common difference is 6 (twice the coefficient of n²).
Tip: If Sₙ = an² + bn, then the common difference is 2a and the first term is a + b.
Sₙ − Sₙ₋₁ = 3n² + 5n − [3(n − 1)² + 5(n − 1)] = 3(2n − 1) + 5 = 6n + 2.
T₁₅ = 6 × 15 + 2 = 92.
Check: T₁ = S₁ = 8 = 6(1) + 2 ✓; the common difference is 6 (twice the coefficient of n²).
Tip: If Sₙ = an² + bn, then the common difference is 2a and the first term is a + b.
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