Question

The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is $1$. Find the common ratio and the terms.

Answer 1

Let $\frac{a}{r},a,ar$ be the first terms of the G.P.
$\frac{a}{r}+a+ar=\frac{39}{10}....\left(1\right)\left(\frac{a}{r}\right)\left(a\right)\left(ar\right)=1...\left(2\right)$

From (2) we obtain $a^3=1\Rightarrow a=1$ (considering real roots only)

Substituting $a=1$ in equation (1), we obtain

$\frac{1}{r}+1+r=\frac{39}{10}$

$\Rightarrow 1+r+r^2=\frac{39}{10}r$

$\Rightarrow 10+10r+{10r}^2-39r=0$

$\Rightarrow {10r}^2-29r+10=0$

$\Rightarrow {10r}^2-25r-4r+10=0$

$\Rightarrow 5r(2r-5)-2(2r-5)=0$

$\Rightarrow (5r-2)(2r-5)=0$

$\Rightarrow r=\frac{2}{5}or\frac{5}{2}$

corresponding terms of the G.P

1. when $r=\frac{2}{5}$

$\Rightarrow \frac{5}{2},1,\frac{2}{5}$

2. when $r=\frac{5}{2}$

$\Rightarrow \frac{2}{5},1,\frac{5}{2}$