Question

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

Answer 1

Let the first three terms of G.P. are $\frac{a}{r},a,ar$

According to question,

$\Rightarrow \frac{a}{r}+a+ar=\frac{13}{12}$ ...... (i)

and $\frac{a}{r}\times a\times ar=-1$

$\Rightarrow a^3=-1$

$\Rightarrow a=-1$

Put $a=-1$ in equation (i),

$\frac{-1}{r}+(-1)-r=\frac{13}{12}$

$\Rightarrow -1-r-r^2=\frac{13}{12}r$

$\Rightarrow -12-12r-12r^2=13r$

$\Rightarrow 12r^2+12r+13r+12=0$

$\Rightarrow 12r^2+25r+12=0$

$\Rightarrow 12r^2+16r+9r+12=0$

$\Rightarrow 4r(3r+4)+3(3r+4)=0$

$\Rightarrow (4r+3)(3r+4)=0$

$r=\frac{-3}{4},\frac{-4}{3}$

Hence, the G.P. for $a=-1$ and

$r=\frac{-3}{4}is\frac{4}{3},-1and\frac{3}{4}$

And the G.P. for $a=-1$ and

$r=\frac{-4}{3}is\frac{3}{4},-1$ and $\frac{4}{3}$