MCQ
The sum of first $40$ positive integers divisible by $6$ is :
- A$2460$
- B$3640$
- ✓$4920$
- D$4860$
✓
Answer
Correct option: C.
$4920$
The frist $40$ positive numbers divisible by $6$ will be
$\{6, 12, 18, 24, ......\}$
Here,
$a = 6$
$d = 6$
$n = 40$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow\text{S}_{40}=\frac{40}{2}\big[2(6)+39(6)\big]$
$\Rightarrow\text{S}_{40}=20[12+234]$
$\Rightarrow\text{S}_{40}=20[246]$
$\Rightarrow\text{S}_{40}=4920$
$\{6, 12, 18, 24, ......\}$
Here,
$a = 6$
$d = 6$
$n = 40$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow\text{S}_{40}=\frac{40}{2}\big[2(6)+39(6)\big]$
$\Rightarrow\text{S}_{40}=20[12+234]$
$\Rightarrow\text{S}_{40}=20[246]$
$\Rightarrow\text{S}_{40}=4920$
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