Respuesta :

Ryan2
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:

[tex]a_n=a_1+(n-1).r\\ \\ 99=3+3(n-1)\\ \\ 99-3=3n-3\\ \\ 3n=99\\ \\ n=33[/tex]

Calculating sum:

[tex]S_{33}=\frac{33(a_1+a_{33})}{2}\\ \\ S_{33}=\frac{33(3+99)}{2}=\frac{33*102}{2}=1683[/tex]
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