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 [tex]n[/tex]-th term is
[tex]3 + 6(n-1) = 6n - 3[/tex]
If there are 24 terms in the sum, then the last term is 6×23 - 3 = 135.
Let [tex]S[/tex] be the sum,
[tex]S = 3 + 9 + 15 + \cdots + 123 + 129 + 135[/tex]
Reverse the order of terms:
[tex]S = 135 + 129 + 123 + \cdots + 15 + 9 + 3[/tex]
If we add up the terms in the same positions, we get twice [tex]S[/tex] on the left side, while on the right side we observe that each pair of terms will sum to 138.
[tex]S + S = (3 + 135) + (9 + 129) + (15 + 123) + \cdots + (135 + 3)[/tex]
[tex]2S = 138 + 138 + 138 + \cdots + 138[/tex]
and since there are 24 terms in the sum, the right side is the sum of 24 copies of 138. In other words,
[tex]2S = 24 \times 138[/tex]
and solving for [tex]S[/tex] gives
[tex]S = \dfrac{24\times138}2 = \boxed{1656}[/tex]
