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

Solution:

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,

$999=3+(n-1) 6$

$999=3+6 n-6$

 

$999=6 n-3$

$999+3=6 n$

Further simplifying,

$1002=6 n$

$n=\frac{1002}{6}$

$n=167$

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

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

For n = 167, we get,

$S_{n}=\frac{167}{2}[2(3)+(167-1) 6]$

$=\frac{167}{2}[6+(166) 6]$

$=\frac{167}{2}(6+996)$

 

$=\frac{167}{2}(1002)$

On further simplification, we get,

$S_{x}=167(501)$

$=83667$

Therefore, the sum of all the odd numbers lying between 1 and 1000 is $S_{n}=83667$.

 

Hence proved