The sum of the 2nd and 3rd term is 12, and the sum of the 3rd term and 4th term is 60. This is the 9th question in my assignment. It's another confusing question.
Here's the question:
In a geometric progression, the sum of the 2nd and 3rd terms is 12, and the sum of the 3rd and 4th term is 60. Find the common ratio of the first term.

Here's what I did:


I stopped there.
 A: You correctly found that 
\begin{alignat*}{3}
ar & + & ar^2 & = 12 \tag{1}\\
ar^2 & + & ar^3 & = 60 \tag{2}
\end{alignat*}
If we multiply equation 1 by $r$ and subtract it from equation 2, we obtain 
$$0 = 60 - 12r \tag{3}$$
Solving equation 3 for $r$ yields the value $r = 5$.  If you also need to find the value of the initial term, substitute $5$ for $r$ in equation 1, then solve for $a$.
A: So you have that 
$ 
\begin{align*}
ar + ar^2 = 12 \\
ar^2 + ar^3 = 60
\end{align*}
$
Then 
\begin{align*}
a = \frac{12}{r + r^2} \\
ar^2 + ar^3 = 60
\end{align*}
$\implies$
\begin{align*}
a = \frac{12}{r + r^2} \\
\frac{12}{r + r^2}r^2 + \frac{12}{r + r^2}r^3 = 60
\end{align*}
$\implies$
\begin{align*}
\frac{12}{r + r^2}r^2 + \frac{12}{r + r^2}r^3 = 60\frac{12}{r + r^2}\frac{r + r^2}{12}
\end{align*}
$\implies$
\begin{align*}
12r^2 + 12r^3 = 60(r + r^2)
\end{align*}
$\implies$
\begin{align*}
12r^3 - 48r^2 - 60r = 0
\end{align*}
$\implies$
\begin{align*}
12r^2 - 48r - 60 = 0
\end{align*}
$
Now solve this equation and you will find 2 solutions. Then you should verify which one satisfies the first equation.
A: Hint take $ar$ common from first equation and $ar^2$ common from second equation and then divide the two you will get $r=\text{common ratio}=5$
A: We have a set of simultaneous equations, $ar     + ar^2 = 12, ar^2 + ar^3=60.$
Divide the second equation by $r$: $ar + ar^2 = \frac{60}{r}$
If $ar     + ar^2 = 12$ and $ar + ar^2 = \frac{60}{r},$ then $\frac{60}{r} = 12.$ Solving for $r$:
$$12r = 60 \to r = 5$$
Now to find $a$, substitute $r$ in any of your equations.
$$(5)a+(5^2)a=12 \to 30a = 12 \to a = \frac{2}{5}$$
