Question 1181056


the sum of the first six terms of the geometric sequence: 3, 12, 48

find{{{ r}}}

{{{ r=12/3=4}}}

For {{{r}}}≠{{{1}}}, the sum of the first n terms of a geometric series is given by the formula 

{{{s=a(1-r^n)/(1-r)}}}

if {{{a=3}}}, {{{r=4}}}, and {{{n=6}}}

{{{s=3(1-4^6)/(1-4)}}}

{{{s=3(-4095)/(-3)}}}

{{{s=(-4095)/(-1)}}}

{{{s=4095}}}


answer: c. {{{4095}}}