#### Answer

Explained below.

#### Work Step by Step

Here, we have $ a_1=1$ and $ d=1$
The general formula to find the sum of the first n terms of a Geometric sequence is given as: $ S_{n}=\dfrac{n(a_1+a_n)}{2}$
Now, $ S_{n}=\dfrac{n(1+n)}{2}+1+2+3+......+n $
and $ S_n=n+(n-1) +......+1$
$ S_n+S_n=(n+1)+(n+1)+......+(n+1)$
$\implies S_n=\dfrac{n(n+1)}{2}$
Hence, $ S_3=1+2+3=\dfrac{3(3+1)}{2}=6$