Question

Find the sum of all integers between 100 and 550, which are divisible by 9.

Answer 1

In this problem, we need to find the sum of all the multiples of 9 lying between 100 and 550.

So, we know that the first multiple of 9 after 100 is 108 and the last multiple of 9 before 550 is 549.

Also, all these terms will form an A.P. with the common difference of 9.

So here,

First term (a) = 108

Last term (l) = 549

Common difference (d) = 9

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

`549 = 108 + (n - 1)9

549 = 108 + 9n - 9

549 = 99 + 9n

Further simplifying,

450 = 9n

`n= 450/9`

n = 50

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

`S_n  =n/2[2a + (n -1)d]`

We get

`S_n = 50/2 [2(108) + (50 - 1)d]`

= 25[216 + (49)9]

= 25(216 + 441)

= 25(657)

= 16425

Therefore, the sum of all the multiples of 9 lying between 100 and 550 is `S_n= 16425`