Question

The sum of of first three terms of a G.P. is $\frac{13}{12}$ and their products is $-1$. Find the common ratio and the terms.

Answer 1

Step $1:$ Solving for value of $r$.
Let the three terms in G.P. be $\frac{a}{r},a,ar$
$\because$ sum of first three terms of G.P. $=\frac{13}{12}$
$\therefore \frac{a}{r}+a+ar=\frac{13}{12}\cdots \left(i\right)$

Also, product of first three terms of G.P. $=-1$
$\Rightarrow \frac{a}{r}\times a\times ar=-1$
$\Rightarrow a^3=(-1{)}^3$
$\therefore a=-1$
Substituting value of $a=-1$ in equation $\left(i\right)$
$-\frac{1}{r}+(-1)+(-1)r=\frac{13}{12}$
$\Rightarrow -\frac{1}{r}-1-r=\frac{13}{12}$
$\Rightarrow \frac{1+r+r^2}{r}=-\frac{13}{12}$
$\Rightarrow 12(1+r+r^2)=-13r$
$\Rightarrow 12+12r+12r^2+13r=0$
$\Rightarrow 12r^2+25r+12=0$

Step $2:$ Solving quadratic equation by factorization method.
$12r^2+25r+12=0$
$\Rightarrow 12r^2+16r+9r+12=0$
$\Rightarrow 4r(3r+4)+3(3r+4)=0$
$\Rightarrow (3r+4)(4r+3)=0$
$\therefore r=-\frac{3}{4}$ or $-\frac{4}{3}$

Step $3:$ Finding required first three terms.

Case $1:$ When $a=-1$ and $r=-\frac{3}{4}$
First term of G.P $=\frac{a}{r}=\frac{-1}{-\frac{3}{4}}=\frac{4}{3}$
Second term of G.P. $=a=-1$
Third term of G.P. $=ar=-1\times -\frac{3}{4}=\frac{3}{4}$

Case $2:$ When $a=-1$ and $r=-\frac{4}{3}$
First term of G.P. $=\frac{a}{r}=\frac{-1}{-\frac{4}{3}}=\frac{3}{4}$
Second term of G.P. $=a=-1$
Third term of G.P. $=ar=-1\times -\frac{4}{3}=\frac{4}{3}$