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 the first three in G.P be $a,ar,a{r}^2$

product of three terms$=\left(a\right).\left(ar\right).\left(a{r}^2\right)=1$

$(ar{)}^3=1$

$a=\frac{1}{r}$

sum of the first three terms$=a+ar+a{r}^2=\frac{39}{10}$

$a(1+r+r^2)=\frac{39}{10}$

$\left(\frac{1}{r}\right)(1+r+r^2)=\frac{39}{10}$

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

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

$10r^2-25r-4r+10=0$

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

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

$4r=2/5$ or $r=5/2$

we take $r=2/5$

first term is$=5/2$

second term is$=\left(5/2\right)\left(2/5\right)=1$

third term is$=\left(5/2\right)(2/5{)}^2=2/5$