Question 1128311
WE are given:
{{{b[1]=48}}}= the first term
{{{b[3]=12}}}= the third term
 
For a G.P. (geometric progression, geometric sequence in the USA),
with a common ratio {{{r}}} , the {{{n^th}}} term is
{{{b[n]=b[1]*r^(n-1)}}}
The third term is given by that "formula" for {{{n=3}}} as
{{{b[3]=b[1]*r^(3-1)=b[1]*r^2}}}
Substituting into {{{b[3]=b[1]*r^2}}}
{{{48}}} for {{{b[1]}}} and {{{12}}} for {{{b[3]}}} we get
{{{12=48*r^2}}}
{{{12/48=r^2}}}
{{{1/4=r^2}}}
Without more information, we cannot tell
if {{{r=1/2=0.5}}} or {{{r=-1/2=-0.5}}}
 
For a G.P. the sum of the first {{{n}}} terms is given by
{{{S[n]=b[1](r^n-1)/(r-1)}}}
 
For {{{n=11}}} we get
{{{S[11]=48(r^11-1)/(r-1)}}}
 
IF {{{r=0.5}}} :
{{{S[11]=48(0.5^11-1)/(0.5-1)}}}
{{{S[11]=48(0.000244140625-1)/(-0.5)}}}
{{{S[11]=48(-0.999755859375)/(-0.5)}}}
{{{S[11]=48*0.4998779296875}}}
{{{S[11]=95.953125}}}
 
IF {{{r=-0.5}}} :
{{{S[11]=48((-0.5)^11-1)/(-0.5-1)}}}
......
{{{S[11]=48(0.000244140625-1)/(-1.5)}}}
...............
{{{S[11]=32.015625}}}