The sum of first 24 terms of the list of number whose $n^{\text{th}}$  term is given by $a_n=3+2n$

So, $a_1=3+2=5$

$a_2=3+2\times 2=7$

$a_3=3+2\times 3=9$

$⋮$

List of numbers becomes 5,7,9,11,…

Here, $7-5=9-7=11-9=2$  and so on.

So, it forms an A.P. with common difference $d=2$

To find $S_{24}$ , we have $n=24,a=5,d=2$

Therefore, $S_{24}=\frac{24}{2}\left[2\times 5+\left(24-1\right)\times 2\right]=12\left(10+49\right)=672$

So sum of first 24 terms of the list of numbers is 672.

(OR)

$S_n=\Sigma a_n$

            $=\Sigma \left(3+2n\right)$

            $=3n+n\left(n+1\right)$

            $=n^2+4n$

$S_{24}={\left(24\right)}^2+4\left(24\right)$

            $=576+96$

            $=672$