Question

What is the sum of the arithmetic sequence 3, 9, 15..., if there are 24 terms? (5 points)

Answer 1

There's a fixed difference of 6 between terms (9 - 3 = 6, 15 - 9 = 6, and so on). The first term of the sequence is 3, so the $n$-th term is

$3 + 6(n-1) = 6n - 3$

If there are 24 terms in the sum, then the last term is 6×23 - 3 = 135.

Let $S$ be the sum,

$S = 3 + 9 + 15 + \cdots + 123 + 129 + 135$

Reverse the order of terms:

$S = 135 + 129 + 123 + \cdots + 15 + 9 + 3$

If we add up the terms in the same positions, we get twice $S$ on the left side, while on the right side we observe that each pair of terms will sum to 138.

$S + S = (3 + 135) + (9 + 129) + (15 + 123) + \cdots + (135 + 3)$

$2S = 138 + 138 + 138 + \cdots + 138$

and since there are 24 terms in the sum, the right side is the sum of 24 copies of 138. In other words,

$2S = 24 \times 138$

and solving for $S$ gives

$S = \dfrac{24\times138}2 = \boxed{1656}$