#### Question

Find the sum of all integers between 50 and 500, which are divisible by 7.

#### 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+ 7n - 7

497 = 49 +7n

497 - 49 = 7n

Further simplifying

448 = 7n

`n = 448/7`

n = 64

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

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

For n = 64 we get

`S_n = 64/2 [2(56) + (64 - 1) 7]`

= 32 [112 + (63)7]

= 32(1123 + 441)

= 32(112 + 441)

= 32(553)

= 17696

Therefore the sum of all the multiples of 7 lying betwenn 50 and 500 is `S_n = 17696`