Question

If the product of three consecutive terms is G.P. is $216$ and sum of their products is pairs is $156$, find them

Answer 1

Let $\frac{a}{r},a,ar$ be the first three terms of an G.P.

It is given that the product of the terms is $216$ that is:

$\frac{a}{r}\times a\times ar=216\Rightarrow a^3=216\Rightarrow a^3=6^3\Rightarrow a=6$

It is also given that the sum of their products is $156$ that is:

$(\frac{a}{r}\times a)+(a\times ar)+(ar\times \frac{a}{r})=156\Rightarrow \frac{a^2}{r}+a^{2}r+a^2=156\Rightarrow a^2(\frac{1}{r}+r+1)=156\Rightarrow a^2\left(\frac{1+r^2+r}{r}\right)=156\Rightarrow a^2\left(\frac{r^2+r+1}{r}\right)=156$

Substituting $a=6$, we get,

$6^2\left(\frac{r^2+r+1}{r}\right)=156\Rightarrow 36\left(\frac{r^2+r+1}{r}\right)=156\Rightarrow \frac{r^2+r+1}{r}=\frac{156}{36}\Rightarrow \frac{r^2+r+1}{r}=\frac{13}{3}$

$\Rightarrow 3(r^2+r+1)=13r\Rightarrow 3r^2+3r+3=13r\Rightarrow 3r^2+3r-13r+3=0\Rightarrow 3r^2-10r+3=0\Rightarrow 3r^2-9r-r+3=0$

$\Rightarrow 3r(r-3)-1(r-3)=0\Rightarrow (3r-1)(r-3)=0\Rightarrow 3r-1=0,r-3=0\Rightarrow 3r=1,r=3\Rightarrow r=\frac{1}{3},r=3$

Now, if $a=6$ and $r=\frac{1}{3}$ then the first three terms of the G.P are:

$\frac{a}{r}=\frac{6}{\frac{1}{3}}=6\times 3=18$

$a=6$ and

$ar=6\times \frac{1}{3}=2$

And if $a=6$ and $r=3$ then the first three terms of the G.P are:

$\frac{a}{r}=\frac{6}{3}=2$

$a=6$ and

$ar=6\times 3=18$

Hence, the three terms are $18,6,2$ or $2,6,18$