# 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.

## To find the sum of all consecutive numbers from 101 to 200, you can either use the method of finding the sum of an arithmetic sequence or manually add up all the numbers.

Method 1: Using the formula for the sum of an arithmetic sequence.

The formula for the sum of an arithmetic sequence is S = (n/2)(a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term.

In this case, the first term (a) is 101 and the last term (l) is 200. The number of terms (n) can be calculated as (l - a) + 1.

So, n = (200 - 101) + 1 = 100.

Now, substitute these values into the formula:

S = (100/2)(101 + 200) = 50(301) = 15,050.

Therefore, the sum of all consecutive numbers from 101 to 200 is 15,050.

Method 2: Manually adding up the numbers.

To manually find the sum of all consecutive numbers from 101 to 200, you can add up each pair of numbers starting from the smallest and largest and work your way towards the middle.

Then consider the pair (102, 199) and add them together: 102 + 199 = 301.

Continue this process until you reach the middle pair (150, 151), which also adds up to 301.

Since there are 100 pairs in total, you have 100 instances of the sum 301.

Therefore, the sum of all consecutive numbers from 101 to 200 is also 301 multiplied by the number of pairs, which is 100.

So, the sum is 301 * 100 = 30,100.

Both methods yield the same result, which is 15,050.

## To find the sum of all consecutive numbers from 101 to 200, you can use the formula for the sum of an arithmetic sequence.

The formula is: S = (n/2)(a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term.

In this case, the first term is 101 and the last term is 200. The number of terms can be calculated by subtracting the first term from the last term and adding 1: n = 200 - 101 + 1 = 100.

Plugging these values into the formula, we get: S = (100/2)(101 + 200) = (50)(301) = 15,050.

Therefore, the sum of all consecutive numbers from 101 to 200 is 15,050.