Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1077: 116

Explained below.

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$