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Q : 12    Find the sum of the first \small 40 positive integers divisible by  \small 6.

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best_answer

Positive integers divisible by 6 are 
6,12,18,...
This is an AP with 
here, \ a = 6 \ and \ d = 6
Now, we know that 
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{40}= \frac{40}{2}\left \{ 2\times 6+(40-1)6 \right \}
\Rightarrow S_{40}= 20\left \{12+234 \right \}
\Rightarrow S_{40}= 20\left \{246 \right \}
\Rightarrow S_{40}= 4920
Therefore,  sum of the first \small 40 positive integers divisible by  \small 6 is 4920

Posted by

Gautam harsolia

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