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determine the sum of all natural numbers between 2 and 100 which are exactly divisible by 3
Asked by suneelmsc | 25 Apr, 2020, 08:04: PM
The first natural number between 2 and 100 which is exactly divisible by 3 is 3.
The last natural number between 2 and 100 which is exactly divisible by 3 is 99.
The numbers exactly divisible by 3 between 2 and 100 are 3, 6, 8, 12,..., 99
This forms an A.P. with first term a=3, common difference d=3 and last term an=99
Since, an=99
99=a+(n-1)d
99=3+(n-1)3
96=3n-3
3n=99
n=33
Number of terms is 33.

Sum of these terms = 33/2[2a+(33-1)d]
=33/2[6 + 32 x 3]
=99/2[34]
=99 x 17
=1683
Answered by Renu Varma | 25 Apr, 2020, 08:48: PM

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