The sum of the first three terms of a GP is \(\frac{13}{12}\) and their product is −1. Find the common ratio and the terms.

Let the three terms be \(\frac{a}{r},\,a,\,ar\)

\(\frac{a}{r} × a × ar = −1\)
⇒ \(a^3 = −1\)
⇒ \(a = −1\)

\(\frac{−1}{r} − 1 − r = \frac{13}{12}\)
\(−1 − r − r^2 = \frac{13r}{12}\)
⇒ \(−12 − 12r − 12r^2 = 13r\)
⇒ \(12r^2 + 25r + 12 = 0\)
⇒ \((3r + 4)(4r + 3) = 0\)
⇒ \(r = \frac{−4}{3}\) or \(r = \frac{−3}{4}\)

For \(r = \frac{−4}{3}\):
\(\frac{a}{r} = \frac{−1}{\frac{−4}{3}} = \frac{3}{4}\)
So the terms are: \(\frac{3}{4},\,−1,\,\frac{4}{3}\)

For \(r = \frac{−3}{4}\):
\(\frac{a}{r} = \frac{−1}{\frac{−3}{4}} = \frac{4}{3}\)
So the terms are: \(\frac{4}{3},\,−1,\,\frac{3}{4}\)