Odd and even consecutive sums

1 + 3 + 5 ..... 99 = 2500 (eq1)

2 + 4 + 6 .....100 = x (eq2)

(Eq2) - (Eq1) gives

(2-1) + (4-3) + (6-5)....(100-99) = x - 2500

Therefore, 50(1) = x - 2500

So, x = 2550

summation of (1+2+3+.......+100)= $\frac {n(n+1)}{2}$

or, the summation is, = $\frac {100(100+1)}{2}$ = $5050$

now, (1+3+5+....99)=2500

so, (2+4+6+........100)= $5050 -2500$ = $2550$

It is interesting to observe that when we add these two equations, we are going to get the sum of numbers from $1$ to $100$ which is $\dfrac{100\times101}{2}=5050$ .

Now, as the question says to find $2+4+6+\cdots +100$ , we simply subtract the sum of $1+3+5+\cdots+ 99$ which is $2500$ (given) from $5050$ to obtain $\boxed{2550}$ as our result.

The sequence of numbers is an arithmetic progression .

$n=\dfrac{100}{2}=50$

$S=\dfrac{50}{2}(2+100)=$ $\large \color{#D61F06}\boxed{2550}$

The difference is 50

1+2+3+........+99 = 4950, but in Q. given is = 2500
So 1 is = 0.505050505051 & 100 is = 50.5050505051
Therefore 2+3+4+....................+100 = 2550 [ 2500 - 0.50505051 + 50.50505051 = 2550