Question

Find $3$ terms in G.P. whose sum and product respectively are $21$ and $216$

Answer 1

Let the three terms be $\frac{a}{r},a,ar$.

It is given that the product of the terms is $216$, therefore,

$\frac{a}{r}\times a\times ar=216\Rightarrow a^3=216\Rightarrow a^3=6^3\Rightarrow a=6.....\left(1\right)$

It is also given that the sum of the terms is $21$, therefore using equation 1 we have,

$\frac{a}{r}+a+ar=21\Rightarrow \frac{6}{r}+6+6r=21\Rightarrow \frac{6}{r}+6r=21-6\Rightarrow 6(\frac{1}{r}+r)=15\Rightarrow \frac{1+r^2}{r}=\frac{15}{6}=\frac{5}{2}\Rightarrow 2r^2+2=5r\Rightarrow 2r^2-5r+2=0\Rightarrow 2r^2-4r-r+2=0\Rightarrow 2r(r-2)-1(r-2)=0\Rightarrow (2r-1)(r-2)=0\Rightarrow (2r-1)=0,(r-2)=0\Rightarrow r=\frac{1}{2},r=2$

With $a=6$ and $r=\frac{1}{2}$, the three terms of G.P are as follows:

$\frac{a}{r}=\frac{6}{\frac{1}{2}}=6\times 2=12$

$a=6$

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

Thus, the terms are $12,6$ and $3$.

With $a=6$ and $r=2$, the three terms are:

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

$a=6$

$ar=6\times 2=12$

Thus, the terms are $3,6$ and $12$.