Question 1083648
Let's  define a couple of variables:
{{{b}}}= the first of the three consecutive terms in the G.P.
{{{r}}}= the common ratio of the G.P.
Then, the next two (consecutive) terms in the G.P are
{{{b*r}}}= the second of the three consecutive terms in the G.P. ,
and {{{b*r^2}}}= the third of the three consecutive terms in the G.P.
The problem probably meant to tell us that
{{{b}}}= the second term of the A.P.
{{{b*r}}}= the third term of the A.P.
That makes {{{b*r-b}}} the common difference of the A.P.
The  problem probably meant to tell us that
{{{b*r^2}}}= the sixth term of the A.P.
That would make it {{{4}}} common differences more than the second term of the A.P.:
{{{b*r^2=b+4*(b*r-b)}}}
We can simplify and solve for {{{r}}} the equation above
{{{b*r^2=b+4*(b*r-b)}}}
{{{b*r^2=b+4b*r-4b)}}}
{{{b*r^2-4b*r=-3b}}}
{{{b*r^2-4b*r+3b=0}}}
{{{b(r^2-4r+3)=0}}}
{{{b(r-1)(r-3)=0}}}
We have 3 possible options:
1) {{{b=0}}} and {{{r}}} is undefined, 
probably not the answer expected.
All the terms of the A.P. and the G.P. would be zero.
2) {{{r=1}}} , also probably not the answer expected.
All the terms of the A.P. and the G.P. would be {{{b}}}, whatever that value is.
3) {{{highlight(r=3)}}} is probably the expected answer.