Question

Find $3$ terms in G.P. whose sum and product respectively are $19$ 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 $19$, therefore using equation 1 we have,

$\frac{a}{r}+a+ar=19\Rightarrow \frac{6}{r}+6+6r=19\Rightarrow \frac{6}{r}+6r=19-6\Rightarrow 6(\frac{1}{r}+r)=13\Rightarrow \frac{1+r^2}{r}=\frac{13}{6}\Rightarrow 6r^2+6=13r\Rightarrow 6r^2-13r+6=0\Rightarrow 6r^2-9r-4r+6=0\Rightarrow 3r(2r-3)-2(2r-3)=0\Rightarrow (2r-3)(3r-2)=0\Rightarrow (2r-3)=0,(3r-2)=0\Rightarrow r=\frac{3}{2},r=\frac{2}{3}$

With $a=6$ and $r=\frac{3}{2}$, the three terms of G.P are as follows:

$\frac{a}{r}=\frac{6}{\frac{3}{2}}=6\times \frac{2}{3}=2\times 2=4$

$a=6$

$ar=6\times \frac{3}{2}=3\times 3=9$

Thus, the terms are $4,6$ and $9$.

With $a=6$ and $r=\frac{2}{3}$, the three terms are:

$\frac{a}{r}=\frac{6}{\frac{2}{3}}=6\times \frac{3}{2}=3\times 3=9$

$a=6$

$ar=6\times \frac{2}{3}=2\times 2=4$

Thus, the terms are $9,6$ and $4$.