<s>[INST] 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.
 [/INST] 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.
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