# Show that the sum of all odd integers between 1 and 1000

which are divisible by 3 is 83667.

Last updated date: 25th Mar 2023

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Answer

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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.

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.

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