What is the sum of the integers from -5001 to 5003, inclusive?

Question

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Answer 1

Sum of n terms = (n/2)(a+l) -----------------(1)

Where n = number of terms

a = first term

d = difference between two consecutive terms.

l = last term

So here,

a = -5001

d = 1

l = 5003

Also, last_term, l = a + (n-1)d

So, n = (l - a)/d +1

=> n = ( 5003 - (-5001))/1 +1

=> n = 10004 + 1 = 10005

Therefore using identity (1) we get

sum = (-5001 + 5003)* (10005)/2

=> sum = 2 * (10005)/2

=> sum = 10005

Answer 2

Let’s write down the series:

-5001, -5000, -4999, ..., -1, 0, 1, ..., 4999, 5000, 5001, 5002, 5003.

The negatives and positives cancel and we’re left with 5002+5003=10005.