Question

The sum of first 3 terms of GP is 13/12 and their product is -1. Find the common ratio and the terms.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{first}3\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{GP}\mathrm{be}\frac{a}{r},a\mathrm{and}ar. \\ \mathrm{Now},\mathrm{sum}\mathrm{of}3\mathrm{terms}=\frac{13}{12} \\ \Rightarrow \frac{a}{r}+a+ar=\frac{13}{12}.....\left(1\right) \\ \mathrm{Product}\mathrm{of}3\mathrm{terms}=-1 \\ \Rightarrow \frac{a}{r}\times a\times ar=-1 \\ \Rightarrow a^3=-1 \\ \Rightarrow a=-1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}a\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ -\frac{1}{r}-1-r=\frac{13}{12} \\ \Rightarrow r+\frac{1}{r}+\frac{25}{12}=0 \\ \Rightarrow 12r^2+25r+12=0 \\ \Rightarrow 12r^2+16r+9r+12=0 \\ \Rightarrow 4r\left(3r+4\right)+3\left(3r+4\right)=0 \\ \Rightarrow \left(4r+3\right)\left(3r+4\right)=0 \\ \Rightarrow 4r+3=0\mathrm{or}3r+4=0 \\ \Rightarrow r=-\frac{3}{4}\mathrm{or}r=-\frac{4}{3} \\ \mathrm{So},\mathrm{common}\mathrm{ratio},r=-\frac{3}{4}or-\frac{4}{3}. \\ \mathrm{When}r=-\frac{3}{4},\mathrm{then}\mathrm{terms}\mathrm{are}: \\ \frac{a}{r}=-1\times \left(-\frac{4}{3}\right)=\frac{4}{3} \\ a=-1 \\ ar=-1\times \left(-\frac{3}{4}\right)=\frac{3}{4} \\ \mathrm{When}r=-\frac{4}{3},\mathrm{then}\mathrm{terms}\mathrm{are}: \\ \frac{a}{r}=-1\times \left(-\frac{3}{4}\right)=\frac{3}{4} \\ a=-1 \\ ar=-1\times \left(-\frac{4}{3}\right)=\frac{4}{3} \end{gathered}$$