Find the sum of the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

Given:

First 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

To do:

We have to find the sum of the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

Solution:

(a) First 40 positive integers divisible by 3 are \( 3,6,9,12,15, \ldots, 120 \)

Here,

\( a=3, d=3 \) and \( l=120 \)
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\mathrm{S}_{40}=\frac{40}{2}[2 \times 3+(40-1) \times 3]$
$=20[6+39 \times 3]$
$=20[6+117]$

$=20 \times 123$

$=2460$

(b) First 40 multiple of 5 are \( 5,10,15,25, \ldots, 200 \)

Here,

\( (a)=5 \), \( (d)=5 \) and \( (l)=200 \)
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\mathrm{S}_{40}=\frac{40}{2}[2 \times 5+(40-1) \times 5]$
$=20[10+39 \times 5]$

$=20[10+195]$
$=20 \times 205$

$=4100$
(c) First 40 multiples of 6 are \( 6,12,18,24, \ldots, 240 \)

Here,

\( (a)=6 \), \( (d)=6 \) and $l=240$
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\mathrm{S}_{40}=\frac{40}{2}[2 \times 6+(40-1) \times 6]$
$=20[12+39 \times 6]$

$=20[12+234]$
$=20 \times 246$

$=4920$

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Updated on: 10-Oct-2022

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