# The sum of first $3$ terms of a G.P is $\large\frac{39}{10}$ and their product is $1$. Find the common ratio and the terms of the G.P.

Toolbox:
• Assume any three terms of a G.P as $\large\frac{a}{r},$$a,ar • Sum of n terms of a G.P.=S_n=a.\large\frac{r^n-1}{r-1} Let the three terms of the G.P. be \large\frac{a}{r},$$a,ar$
Given that sum of the three terms$=\large\frac{39}{10}$
$\Rightarrow\:\large\frac{a}{r}$$+a+ar=\large\frac{39}{10}.......(i) Also given that their product =1 \Rightarrow\:\large\frac{a}{r}$$\times a \times ar=1$
$\Rightarrow\:a^3=1$ or $a=1$
Substituting the value of $a$ in (i) we get
$\large\frac{1}{r}$$+1+r=\large\frac{39}{10} \Rightarrow\:\large\frac{1+r+r^2}{r}=\frac{39}{10} \Rightarrow\:10r^2+10r+10=39r \Rightarrow\:10r^2-29r+10=0 Solving this quadratic equation we get \Rightarrow\:10r^2-25r-4r+10=0 \Rightarrow\:5r(2r-5)-2(2r-5)=0 \Rightarrow\:(2r-5)(5r-2)=0 r=\large\frac{5}{2}$$\:\:or\:\:r=\large\frac{2}{5}$
Substituting the value of $a\:\:and\:\:r$ the three numbers are
$\large\frac{2}{5}$$,1,\large\frac{5}{2}$
answered Feb 26, 2014