Question

The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

Answer 1

Let the three numbers in G.P. be $\frac{a}{r},a,ar$

Then product of them is $\left(\frac{a}{r}\right)+\left(a\right)+\left(ar\right)=21$ ............(i)

$=\frac{a}{r}(1+r+r^2)=21$

and sum of their squares

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

Now,

$a(1+r+r^2)=21r$ ......(iii)

Then, $a^2(1+r+r^2{)}^2=441r^2$

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

$189r^2+2ar\times 21r=441r^2$

Dividing both sides by 21 $r^2$

$9+2a=21$

$2a=21-9=12$

$a=6\Rightarrow a=6$

Putting in (iii)

$6(1+r+r^2)=21r$

$6+6r+6r^2-21r=0$

$6r^2-15r+6=0$

$6r^2-12r-3r+6=0$

$\Rightarrow 6r(r-2)-3(r-2)=0$

$\Rightarrow (6r-3)(r-2)=0$

$\therefore r=2,\frac{1}{2}$

Hence, the G.P. for a = 6 and r = 2 is 12, 6 and 3

And the G.P. for a = 6 and

$r=\frac{1}{2}$ is 3, 6 and 12