Question

What is the sum of all integers from -500 to 500?

Answer 1

Hint: We are given a range of numbers more specifically a range of integers and we are asked to find the sum of these numbers. We will use the formula of sum of the integers \[S=\dfrac{n\left( a+l \right)}{2}\]. Here, we have the values of ‘n’ as the number of terms and ‘a’ as the initial term and ‘l’ as the final term. We will first find the values of the variables used in the above formula if we don’t have the values. Then, we will substitute and compute the values and we will have the sum of the given integers.

Complete step-by-step solution:
According to the given question, we are given a range of integers and we are asked to find the sum of the given integers.
The given range of integers is, -500 to 500
We cannot go about by adding each term at a time, because that will be very time consuming.
In order to find the sum of the integers, we will use the following formula,
\[S=\dfrac{n\left( a+l \right)}{2}\]
Here,
‘n’ is the number of terms involved
‘a’ is the first term
‘l’ is the last term
From the given range, we will find the values of all the above required variables.
The first term, \[a=-500\]
The last term, \[l=500\]
Number of terms, \[n=500-(-500)\]
\[n=500+500=1000\]
We will now substitute the values in the formula of the sum of the integers and we get,
\[\Rightarrow S=\dfrac{1000\left( -500+500 \right)}{2}\]
We will calculate further and we get,
\[\Rightarrow S=\dfrac{1000\left( 0 \right)}{2}\]
And so we get,
\[\Rightarrow S=\dfrac{0}{2}=0\]
Therefore, the sum of all the integers in the given range is 0.

Note: In the given question, we were given a range of integers that included both positive and negative numbers and that with the same upper and lower limit. So, whatever the value of either positive or negative numbers adding up is cancelled by the same sum of the other set of numbers with opposite signs. So, while reading the question, make note of the range and numbers involved to get a hint of the answer you will eventually get.