Question

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

Answer 1

Hint: First, check whether the set of required numbers form a series and then find the total number of odd integers divisible by 3 between 1 and 1000. After that make a sum of that using the appropriate sum of series formula.

As, we all know that all odd numbers between 1 and 1000,
which are divisible by 3 are \[{\text{3, 9, 15, }}......{\text{ 999}}\] which forms an A.P.
\[ \Rightarrow \]First term of this A.P is \[{a_1} = 3\].
\[ \Rightarrow \]Second term of this A.P. is \[{a_2} = 9\].
\[ \Rightarrow \]Last term of this A.P. is \[{a_n} = 999\].
\[ \Rightarrow \]Common difference \[d = {a_2} - {a_1} = 9 - 3 = 6\]
So, we know that \[{n^{th}}\] term of any A.P is given as
\[ \Rightarrow {a_n} = {a_1} + (n - 1)d\]
For, finding the value of n.
On putting, \[{a_n} = 999,{\text{ }}{a_1} = 3\] and \[d = 6\] in the above equation. We get,
\[
   \Rightarrow 999 = 3 + (n - 1)6 \\
   \Rightarrow 999 = 3 + 6n - 6 \\
\]
On solving the above equation. We get,
\[ \Rightarrow n = \dfrac{{1002}}{6} = 167\] numbers in the A.P
Now, as we know that sum of these n terms of A.P is given by,
\[ \Rightarrow {S_n} = \dfrac{n}{2}[{a_1} + {a_n}]\]
So, putting values in the above equation. We get,
So, putting values in the above equation. We get,
\[ \Rightarrow {S_{167}} = \dfrac{{167}}{2}[3 + 999] = \dfrac{{167}}{2}*1002 = 167*501 = 83667\]
\[ \Rightarrow \]Hence, the sum of all odd numbers between 1 and 1000 which are divisible by 3 is \[{S_{167}} = 83667\].

Note: Whenever we came up with this type of problem then first, we find value of n using value of \[{{\text{n}}^{th}}\] term formula in an A.P and then, we can easily find sum of n terms of that A.P using formula of sum of n terms of A.P, if first and last term are given.