Question
Find the sum of the first $40$ positive integers divisible by $6$.
✓
Answer
The first $40$ positive integers divisible by $6$ are $6, 12, 18, 24, .....$
Here, $a_2 - a_1 = 12 - 6 = 6$
$a_3 - a_2 = 18 - 12 = 6$
$a_4 - a_3 = 24 - 18 = 6$
i.e.$ a_{k+1} - a_k$ is the same every time.
So, the above list of numbers form an AP.
Here, $a = 6$
$d = 6$
$n = 40$
$\therefore $ Sum of the first 40 positive integers = $S_{40}$
$ = \frac{{40}}{2}\left[ {2a + (40 - 1)d} \right]$ .........${\{\because {S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]}\}$
= 20[2a + 39d]
$ = (20)[2 \times 6 + 39 \times 6]$
$= (20) (246)$
$= 4920$
Here, $a_2 - a_1 = 12 - 6 = 6$
$a_3 - a_2 = 18 - 12 = 6$
$a_4 - a_3 = 24 - 18 = 6$
i.e.$ a_{k+1} - a_k$ is the same every time.
So, the above list of numbers form an AP.
Here, $a = 6$
$d = 6$
$n = 40$
$\therefore $ Sum of the first 40 positive integers = $S_{40}$
$ = \frac{{40}}{2}\left[ {2a + (40 - 1)d} \right]$ .........${\{\because {S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]}\}$
= 20[2a + 39d]
$ = (20)[2 \times 6 + 39 \times 6]$
$= (20) (246)$
$= 4920$
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