Question 980822
In a geometric sequence with first term {{{b[0]}}} and common ratio {{{r}}} ,
{{{S[n]=b[0]((r^n-1)/(r-1))}}}= sum of the first {{{n}}} terms.
{{{S[3]=b[0]((r^3-1)/(r-1))}}}= sum of the first {{{3}}} terms.
{{{S[6]=b[0]((r^6-1)/(r-1))}}}= sum of the first {{{n}}} terms.
The sum of the fourth term to the sixth term is
{{{S[6]-S[3]=b[0]((r^6-1)/(r-1))-b[0]((r^3-1)/(r-1))=b[0]((r^6-r^3)/(r-1))=b[0]r^3((r^3-1)/(r-1))}}} .
For the geometric sequence described in this problem,
{{{b[0]r^3((r^3-1)/(r-1))=(1/3)b[0]((r^3-1)/(r-1))}}}--->{{{r^3=1/3}}}--->{{{r=root(3,1/3)}}} or {{{r=(1/3)^(1/3)}}}


NOTE:
We do not know what the first term is.
There is an infinite number of different geometric sequences (with different first terms) that fit the description of the problem.
All of them have the same common ratio.