Question

The sum of three terms of a geometric sequence 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 three terms of an G.P.

It is given that the sum of the terms is $\frac{39}{10}$ that is:

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

It is also given that the product of the terms is $1$ that is:

$\frac{a}{r}\times a\times ar=1\Rightarrow a^3=1\Rightarrow a^3=1^3\Rightarrow a=1$

Substitute the value of $a$ in equation 1 as follows:

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

$\Rightarrow 10(1+r+r^2)=39r\Rightarrow 10+10r+10r^2=39r\Rightarrow 10+10r-39r+10r^2=0\Rightarrow 10r^2-29r+10=0\Rightarrow 10r^2-4r-25r+10=0$

$\Rightarrow 2r(5r-2)-5(5r-2)=0\Rightarrow 2r-5=0,5r-2=0\Rightarrow 2r=5,5r=2\Rightarrow r=\frac{5}{2},r=\frac{2}{5}$

Now, if $a=1$ and $r=\frac{5}{2}$ then the first three terms of the G.P are:

$\frac{a}{r}=\frac{1}{\frac{5}{2}}=\frac{2}{5}$

$a=1$ and

$ar=1\times \frac{5}{2}=\frac{5}{2}$

And if $a=1$ and $r=\frac{2}{5}$ then the first three terms of the G.P are:

$\frac{a}{r}=\frac{1}{\frac{2}{5}}=\frac{5}{2}$

$a=1$ and

$ar=1\times \frac{2}{5}=\frac{2}{5}$

Hence, the three terms are $\frac{2}{5},1,\frac{5}{2}$ or $\frac{5}{2},1,\frac{2}{5}$