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

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,

$715=85+(n-1) 5$

$715=85+5 n-5$

$715=80+5 n$

$715-80=5 n$

Further simplifying,

$635=5 n$

$n=\frac{635}{5}$

$n=127$

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

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

We get,

$S_{n}=\frac{127}{2}[2(85)+(127-1) 5]$

$=\frac{127}{2}[170+(126) 5]$

$=\frac{127}{2}(170+630)$

$=\frac{127(800)}{2}$

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 $S_{n}=50800$.