Question

In a Geometric Progression (G.P) the product of first five terms is $1$ and the sum of first three terms is $\frac{7}{4}$. Find its common ratio.

Answer 1

Let the $5$ terms in geometric progression be
$\frac{a}{r^2},\frac{a}{r},a,ar,a{r}^2$
Now given that
Product of these numbers $=1$
i.e., $\frac{a}{r^2}\times \frac{a}{r}\times a\times ar\times a{r}^2=1$
$a^5=1$
$a^5=1^5$
$\Rightarrow$ $a=1$
and
sum of first 3 terms $=\frac{7}{4}$
i.e., $\frac{a}{r^2}+\frac{a}{r}+a=\frac{7}{4}$
$\frac{a+ar+a{r}^2}{r^2}=\frac{7}{4}$
$4(a+ar+a{r}^2)=7r^2$
But $a=1$
$\therefore$ $4(1+1r+1r^2)=7r^2$
Or, $7r^2-4r^2-4r-4=0$
$3r^2-4r-4=0$
Solve by factorisation method
$3r^2-4r-4=0$
$3r^2-6r+2r-4=0$
$3r(r-2)+2(r-2)=0$
$(r-2)(3r+2)=0$
$\Rightarrow$ $r-2=0$ or $3r+2=0$
$r=2$ or $3r=-2$
$r=-\frac{2}{3}$
$\therefore$ Common ratio of geometric progression be $=r=2$ or $-\frac{2}{3}$