Progression (arithmetic)

Let the number of terms of the arithmetic progression be $n$ . Then the AP is $\{3, 5, 7, \cdots 2n+1\}$ and its sum:

$\begin{aligned} \sum_{k=1}^n (2k+1) & = 2600 \\ 2 \times \frac {n(n+1)}2 + n & = 2600 \\ n^2 + 2n & = 2600 \\ n^2 + 2n + 1 & = 2601 \\ (n+1)^2 & = 2601 = 51^2 \\ \implies n & = \boxed {50} \end{aligned}$

If we have $N$ terms in this arithmetic progression, then we need to compute:

$\frac{N}{2} \cdot [3 + 3 + 2(N-1)] = 2600$ ;

or $\frac{N}{2} \cdot (2N+4) = 2600;$

$or N^2 + 2N - 2600 = 0;$

or $(N-50)(N+52) = 0;$

or $N = 50, -52.$ Since we require $N > 0$ , the answer is just $\boxed{N=50}.$