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The sum of all odd integers between 1 and 1000 which are divisible by 3 is:-

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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,

an=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,

On further simplification we get

Sn=167(501)

= 83667

Therefore the sum of all the odd numbers lying between 1 and 1000 is Sn=83667

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by (26.4k points)

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

Given :  

Odd integers between 1 and 1000 which are divisible by 3 are  3, 9 ,15 ..........999.

Here, a = 3 , d = 9 - 3 = 6 , an ,(l) = 999

By using the formula ,an = a + (n - 1)d

999 = 3 + (n - 1)6

999 = 3 + 6n - 6

999 = 6n - 3

999 + 3 = 6n

1002 = 6n

n = 1002/6

n = 167

By using the formula ,Sum of nth terms , Sn = n/2 [a + l]

S167 = 167/2 [3 + 999]

S167 = 167/2 × 1002

S167 = 167 (501)  

S167 = 83667

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

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