Find the next $3$ terms in the sequence: $4,8,12,\mathrm{16...}$

The correct option is D: $20,24$ and $28$

The sequence follows:
$4,8,12,\mathrm{16...}$
We can see, the terms are consecutive multiples of $4.$
$4\times 1=4$

$4\times 2=8$

$4\times 3=12$

$4\times 4=16$

$4\times 5=20$

$4\times 6=24$

$4\times 7=28$

So, the next three terms are $20,24,$ and $28.$

Alternate method

The given sequence is $4,8,12,\mathrm{16...}$

$8-4=4$

$12-8=4$

$16-12=4$

so, there is a common difference between these terms of the given sequence.

So, the given sequence is in AP.

Here, first term, $a=4$, and common difference, $d=4$

We can find the $n^{th}$ term of an A.P by the formula

$a_n=a+(n-1)d$

$a_5=4+(5-1)4=4+16=20$

$a_6=4+(6-1)4=4+20=24$

$a_7=4+(7-1)4=4+24=28$

So, the next three terms are $20,24,28$