Find the sum of all integers between 50 and 500,

Solution:

In this problem, we need to find the sum of all the multiples of 7 lying between 50 and 500.

So, we know that the first multiple of 7 after 50 is 56 and the last multiple of 7 before 500 is 497.

Also, all these terms will form an A.P. with the common difference of 7.

So here,

First term (a) = 56

Last term (l) = 497

Common difference (d) = 7

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,

$497=56+(n-1) 7$

$497=56+7 n-7$

 

$497=49+7 n$

$497-49=7 n$

Further simplifying,

$448=7 n$

$n=\frac{448}{7}$

$n=64$

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

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

For n = 64, we get,

$S_{\pi}=\frac{64}{2}[2(56)+(64-1) 7]$

$=32[112+(63) 7]$

$=32(112+441)$

$=32(553)$

 

$=17696$

Therefore, the sum of all the multiples of 7 lying between 50 and 500 is $S_{n}=17696$.