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Find the sum of the first 40 positive integers divisible by 6.
Given:
First 40 positive integers divisible by 6.
To do:
We have to find the sum of the first 40 positive integers divisible by 6.
Solution:
First 40 multiples of 6 are $6,12,18,24, \ldots, 240$
Here,
$a=6, d=6$ and $l=240$
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\mathrm{S}_{40}=\frac{40}{2}[2 \times 6+(40-1) \times 6]$
$=20[12+39 \times 6]$
$=20[12+234]$ $=20 \times 246$
$=4920$
The sum of the first 40 positive integers divisible by 6 is 4920.
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