How to find fourth term a of geometric series using sum of first three terms and second term?

Question

I need help with this...

Exercise: the second term of a geometric series is 6 and the sum of the first three terms of this same series is -14. Find the value of the fourth term.

This is what I've tried:

First equation:

$6=u_1r^1$

Second equation:

$-14=\frac{u_1(r^3-1)}{r-1}$

$-14=u_1(r^2+r+1)$

Both together:

$\frac{6}{r}=\frac{-14}{r^2+r+1}$

But I keep getting stuck isolating r. Am I doing this right? What if I get two values for r?

Answer 1

From $u_1r=6$ and $u_1(1+r+r^2)=-14$ you can see that $r\ne 0$, and $1+r+r^2\ne 0$. Then you multiply both sides of $$\frac{6}{r}=\frac{-14}{r^2+r+1}$$ by $r(1+r+r^2)$. Solve the quadratic equation. It's perfectly fine if you get two values of $r$.

Answer 2

You should expect two values for $r$ since you only know the middle term of the first three and their sum, and these are unchanged if you reverse the three terms of the progression and change from $r$ to $\frac 1r$.

But this does suggest that there may be two rather different answers for the value of the fourth term.

Solve your quadratic and find them.