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The sum of the first $n$ terms of an arithmetic progression is Sₙ = $3 n^{2} + 5 n$ . Its 10th term is:

Competition practiceAlgebra

A62
B56
C65
D59

Step-by-Step Solution

aₙ =

S

ₙ −

S

ₙ₋₁ =

(3 n^{2} + 5 n) - [3 (n - 1)^{2} + 5 (n - 1)] = 6 n + 2

. So a₁₀ = 60 + 2 = 62. (Check:

S

₁ = 8 and a₁ = 6 + 2 = 8.)

Answer: A) 62

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