How many terms from the progression 3, 5, 7, 9, ... must be taken in order that their sum will be 2600?

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