Question

Sum of three terms which are in G.P. is 21. Sum of the square of the terms is 189. Find the terms.

Answer 1

Let three numbers in G.P be $a,ar,a{r}^2$

Since the sum of the numbers is $21$

$\therefore a+ar+a{r}^2=21$

$\Rightarrow a(1+r+r^2)=21$ .....$\left(1\right)$

Since the sum of the squares of the numbers is $189$

$\therefore a^2+a^2{r}^2+a^2{r}^4=189$

$\Rightarrow a^2(1+r^2+r^4)=189$ ......$\left(2\right)$

Squaring both sides of $\left(1\right)$, we get

$a^2{(1+r+r^2)}^2=441$ .......$\left(3\right)$

Dividing $\left(3\right)$ by $\left(2\right)$, we get

$\frac{a^2{(1+r+r^2)}^2}{a^2(1+r^2+r^4)}=\frac{441}{189}$

$\Rightarrow \frac{{(1+r+r^2)}^2}{(1+r^2+r^4)}=\frac{21}{9}$

$\Rightarrow \frac{{(1+r+r^2)}^2}{(1+2r^2+r^4-r^2)}=\frac{7}{3}$

$\Rightarrow \frac{{(1+r+r^2)}^2}{{(r^2+1)}^2-r^2}=\frac{7}{3}$

$\Rightarrow \frac{{(1+r+r^2)}^2}{(r^2+1-r)(r^2+1+r)}=\frac{7}{3}$

$\Rightarrow \frac{(1+r+r^2)}{(r^2+1-r)}=\frac{7}{3}$

$\Rightarrow 3r^2+3r+3=7r^2-7r+7$

$\Rightarrow 4r^2-10r+4=0$

$\Rightarrow 2r^2-5r+2=0$

$\Rightarrow r=\frac{5\pm \sqrt{5^2-4\times 2\times 2}}{2\times 2}$

$=\frac{5\pm \sqrt{25-16}}{4}$

$=\frac{5\pm \sqrt{9}}{4}=\frac{5\pm 3}{4}=\frac{8}{4},\frac{2}{4}=2,\frac{1}{2}$

When $r=2$ from $\left(1\right)$

$a(1+2+2^2)=21\Rightarrow a=3$

$\therefore$ Numbers are $3,3\times 2,3\times 2^2$ or $3,6,12$

When $r=\frac{1}{2}$ from $\left(1\right)$

$a[1+\frac{1}{2}+\frac{1}{4}]=21$

$\Rightarrow a\left[\frac{4+2+1}{4}\right]=21$

$\Rightarrow a=\frac{21\times 4}{7}=3\times 4=12$

$\therefore$ Numbers are $12,12\times \frac{1}{2},12\times {\left(\frac{1}{2}\right)}^2$ or $12,6,3$