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sum of all multiples of 6 between 6 and 999

Guest Mar 2, 2017
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sum of all multiples of 6 between 6 and 999

 

This is an arithemetic progression.

 

The first term is 6*2=12   and the last term is  6*166=996

 

So that is 166-2+1=165 term

T1=6

n=165

T165= 996

d=6

 

\(Sn=\frac{n}{2}(T_1+Last)\\ S_{165}=\frac{996}{2}(6+996)\\ S_{165}=\frac{996}{2}(1002)\\ \)

 

sum = 996/2*1002 = 498996

Melody  Mar 2, 2017

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