3, 5, 7, 9, ... ← is an AP with
d = 2
$S = \dfrac{n}{2}\left[ 2a_1 + (n - 1)d \right]$
$2600 = \dfrac{n}{2}\left[ 2(3) + (n - 1)(2) \right]$
$2600 = 3n + n(n - 1)$
$n^2 + 2n - 2600 = 0$
$(n - 50)(n + 52) = 0$
$n = 50 ~ \text{and} ~ -52$
Use $n = 50$ ← answer