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Q : 13     Find the sum of the first \small 15  multiples of  \small 8.

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First 15 multiples of 8 are  
8,16,24,...
This is an AP with 
here, \ a = 8 \ and \ d = 8
Now, we know that 
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times 8+(15-1)8 \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 16+112 \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 128 \right \}
\Rightarrow S_{15}= 15 \times 64 = 960
Therefore,  sum of the first 15 multiple of 8 is 960

Posted by

Gautam harsolia

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