Question

Find the sum of all integers between 84 and 719, which are multiples of 5.

Answer 1

In this problem, we need to find the sum of all the multiples of 5 lying between 84 and 719.

So, we know that the first multiple of 5 after 84 is 85 and the last multiple of 5 before 719 is 715.

Also, all these terms will form an A.P. with the common difference of 5.

So here

First term (a) = 85

Last term (l) = 715

Common difference (d) = 5

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now as we know

`a_n = a + (n - 1)d`

So, for the last term,

715 = 85 + (n - 1)5

715 = 85 + 5n - 5

715 = 80 + 5n

Further simplifying 

635 = 5n

`n = 635/5`

n = 127

Now, using the formula for the sum of n terms,

`S_n = n/2 [2a + (n - 1)d]`

We get

`S_n = 127/2 = [2(85) + (127 - 1)5]`

`= 127/2 [170 + (126)5]`

`= 127/2 (170 + 630)`

`= (127(800))/2`

On further simplification, we get,

`S_n = 127(400)`

= 50800

Therefore, the sum of all the multiples of 5 lying between 84 and 719 is `S_n = 50800`