Multiples of 3 from 3 to 99 inclusive is a Aritmetic Progression where:
a1 = 3
an = 99
r = 3
Using the formula of general terms:
\(a_n=a_1+(n-1).r\\
\\
99=3+3(n-1)\\
\\
99-3=3n-3\\
\\
3n=99\\
\\
n=33\)
Calculating sum:
\(S_{33}=\frac{33(a_1+a_{33})}{2}\\
\\
S_{33}=\frac{33(3+99)}{2}=\frac{33*102}{2}=1683\)
