Question

Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

Answer 1

In this problem, we need to prove that the sum of all odd numbers lying between 1 and 1000 which are divisible by 3 is 83667.

So, we know that the first odd number after 1 which is divisible by 3 is 3, the next odd number divisible by 3 is 9 and the last odd number before 1000 is 999.

So, all these terms will form an A.P. 3, 9, 15, 21 … with the common difference of 6

So here

First term (a) = 3

Last term (l) = 999

Common difference (d) = 6

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,

999 = 3 + (n -1)6

999 = 3 + 6n - 6

999 = 6n - 3

999 + 3 = 6n

Further simplifying

1002 = 6n

`n = 1002/6`

n = 167

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

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

For n = 167 we get

`S_n = 167/2 [2(3) + (167 - 1)6]`

`= 167/2 [6 + (166)6]`

`= 167/2 (6 + 996)`

`= 167/2 (1002)`

On further simplification we get

`S_n = 167(501)`

= 83667

Therefore the sum of all the odd numbers lying between 1 and 1000 is `S_n = 83667`

Hence proved