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