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# Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667. Verified
361.8k+ views
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.
Last updated date: 21st Sep 2023
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