Quite complex algebra problem

Find a relation between the two sequences and then solve!

The function consists of two arithmetic progressions: One for the real part, and the other for the imaginary part. First: We divide the function into two different series:

$S1$ = $1 + 3 + 5 + 7 ... x$

$S2$ = $1 + 2 + 3 + 4 ... y$

Second: Since it's a finite arithmetic sequence, we can put an expression to find the sum of all values in "n" terms. (Here N is unknown)

The Expression of the sum of a finite arithmetic sequence is $\frac{n(last + first)}{2}$

So the sum of $S1 =$ $\frac{n(x+1)}{2}$ = $2500$

Then, $n = (\frac{5000}{y+ 1}$ ) (1)

The sum of $S2 =$ $\frac{n(x+1)}{2}$ = $1275$

Third: We substitute the value of n in the expression of $S1$ to get:

$2550x + 2550 = 5000y + 5000$ , (Put (1) instead of n and cross multiplicate) which is

$51x - 100y = 49$ (2)

Fourth: We will find a relation between the first and the second series, which is $2(S2)_k - (S1)_k = 1$ , which leads as to $2y - x = 1$ (3)

Fifth: Now we have two equations for the variables, let's do some math.

Multiply equation (3) by $50$

$100y - 50x = 50$ (4)

Get the sum of equation (4) and (2) $x = 99$

Substitute the value of x in equation (3)

$y = 50$

Sixth: Get the answer $x + y = 149$

so what u need to know is ∑(1 to x) which is equal to 2500; by applying the formula for sum of the series i.e Sn = n/2 (2a+(n-1)d) =2500=n/2 (2(1)+(n-1)2) (since we are doing summation of 1+3+....) =2500=n/2 (2+2n-2) =2500=n/2 (2n) =2500=n^{2} implies that, n=50 so,An=a +(n-1)d = 1+ 49(2) =99 therefore x=99 and since yi is same as n=50 thus, 99+50=149

$\begin{aligned} \sum_{n=1}^y (2n-1 +ni) & = 2500 + 1275 i \\ {\color{#3D99F6}y(y+1) - y} + {\color{#D61F06}\frac {y(y+1)}2}i & = {\color{#3D99F6}2500} + {\color{#D61F06}1275} i & \small \color{#D61F06} \implies \frac {y(y+1)}2 = 1275 \color{#3D99F6} \implies y(y+1) = 2550 \\ \color{#3D99F6} 2550 - y & = \color{#3D99F6} 2500 \\ \implies y & = 50 \\ x & = 2y - 1 = 99 \\ \implies x + y & = 99+50 = \boxed{149} \end{aligned}$