Question 916465
A geometric sequence has a second term of 6:
<pre>
{{{a[2]=6}}}

The first term is the second term divided by the common ratio r.

{{{a[1]=6/r}}}

The third term is the second term multiplied by the common ratio r.

{{{a[3]=6r}}}
</pre>
and the sum of the first  3 term is -14
<pre>
{{{a[1]+a[2]+a[3]=-14}}}

{{{6/r+6+6r=-14}}}

Clear the fraction by multiplying through by r

{{{6+6r+6r^2=-14r}}}

{{{6+20r+6r^2=0}}}

Divide through by 2

{{{3+19r+3r^2=0}}}

Arrange in descending order

{{{3r^2+10r+3=0}}}

Factor:

{{{(3r+1)(r+3)=0}}}

Use zero-factor property:

3r+1=0;   r+3=0
  3r=-1;    r=-3
   r={{{-1/3}}}

So there will be two solutions to this problem.

Using r={{{-1/3}}}, the sequence is

{{{a[1]=6/r=6/(-1/3)=6*(-3/1)=-18}}}

{{{a[2]=6}}}

{{{a[3]=6(-1/3)=-2}}}

{{{a[4]=(-2)(-1/3)=2/3}}}

-18,6,-2,{{{2/3}}}

Fourth term = {{{2/3}}}

-------
Using r = -3, the sequence is

{{{a[1]=6/r=6/(-3)=-2}}}

{{{a[2]=6}}}

{{{a[3]=6(-3)=-18}}}

{{{a[4]=(-18)(-3)=54}}}

-2, 6, -18, 54

Fourth term = 54.

------

Edwin</pre>