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what is the sum of all consecutive numbers from 101 to 200, those two numbers inclusive?

Halfway between, is 150. Then add 149+151,thence 148+152, and so on till 101+199. Each instance you have 300. So it seems to me...

Sum= 200 + 150 + 300*N, where N is the number of pairs you added. You should be able to figure that out.

There is another method, of course: Use the formula for the sum of an arithmetic sequence. It works well here.

The sum of the set of consecutive numbers from "a" to "b" is given by S = a(a + 1)/2 - b(b - 1)/2.

If the smallest number of the group of 50 is 1997, the largest number is 1997 + (50 - 1) = 2046.

Their sum is S = 2046(2047)/2 - 1997(1996)/2 = 101,075 or
...................S = [2046^2 + 2046 - 1997^2 + 1997]/2 = 101,075 or
...................S = (1997 + 2046)(2046 - 1997 + 1)/2 = 101.075.

The equivalent sum of another 25 consecutive numbers, the smallest of which is "x", derives from
(x + 24)(x + 25)/2 - x(x - 1)/2 = 101,075

Expanding, x^2 = 49x + 600 - x^2 + x = 202,150.

Thus, 49x = 210,550 making x = 4031.

The group of 25 conscutive numbers is therefore 4031 - 4055.

Their sum is 4055(4056)/2 - 4031(4030)/2 = 101,075.

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