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 first three term of the given G.P

$\therefore \frac{a}{r}+a+ar=\frac{39}{10}$ .... (1)

and $\frac{a}{r}+a+ar=1$

$\Rightarrow a^3=1\Rightarrow a=1$ ........(2)

Substituting the value of 'a' from (2) in(1),

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

$\Rightarrow 10+10r+10r^2=39r$

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

$\therefore r=\frac{-(-29)\pm \sqrt{(-29{)}^2-4x\times 10\times 10}}{2\times 10}$

$\Rightarrow r=\frac{29\pm \sqrt{841-400}}{20}\Rightarrow r=\frac{29\pm 21}{20}$

Either $r=\frac{29+21}{20}\Rightarrow orr=\frac{29\pm 21}{20}$

$\Rightarrow r=\frac{50}{20}=\frac{5}{2}orr=\frac{8}{20}=\frac{2}{5}$

When $r=\frac{5}{2},$ then the first three terms of the G.P. are

$\frac{1}{\frac{2}{5}},1,1\times \frac{5}{2}i.e.,\frac{5}{2},1,\frac{2}{5}.$

When $r=\frac{2}{5},$ then the first three terms of the G.P. are

$\frac{1}{\frac{2}{5}},1,1\times \frac{2}{5}or\frac{5}{2},1,\frac{2}{5}.$