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

G Gautam harsolia

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

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