Sum of all the natural no. 100 to200 which is mutiply by3
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Your answer is 4950...
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1
a = 102, n=33, d=3
We know that, S = n(2a + (n-1)d)/2
= 33 (2 * 102 + (33-1)*3)/2
= 33 ( 204 + 96)/2
= 4950.
Hope this helps!
We know that, S = n(2a + (n-1)d)/2
= 33 (2 * 102 + (33-1)*3)/2
= 33 ( 204 + 96)/2
= 4950.
Hope this helps!
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