If the first term of an A.P. is 2 and common difference is 4,

In the given problem, we need to find the sum of 40 terms of an arithmetic progression, where we are given the first term and the common difference. So, here we use the following formula for the sum of n terms of an A.P.,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Given,

First term (a) = 2

Common difference (d) = 4

Number of terms (n) = 40

So, using the formula we get,

$S_{40}=\frac{40}{2}[2(2)+(40-1)(4)]$

$=(20)[4+(39)(4)]$

$=(20)[4+156]$

$=(20)(160)$

$=3200$

Therefore, the sum of first 40 terms for the given A.P. is $S_{40}=3200$. So, the correct option is (a).