Geometric progression, two equations problem We have two equations:
$$1. \ a_1 + a_2 + a_3 = 21$$
$$2. \ a_1^2 + a_2^2 + a_3^2 = 189$$
Answer should by $a_1$, $a_2$ and $a_3$.
How the title says, these 3 elements are part of geometric progression.
I have no idea how to solve that problem. Any description on how to solve it mathematically, not just bruteforce it by guessing will be great. Thanks in advance!!
 A: It tells you that these elements are part of a geometric progression, which means that you obtain each successive term after the first by multiplying by a fixed constant $k$, so that $a_2 = ka_1$ and $a_3 = k^2a_1$. This gives us a system of two equations in two variables:
$$a_1 + ka_1 + k^2a_1 = 21$$
$$a_1^2+k^2a_1^2+k^4a_1^2=189$$
Factoring out $a_1$ and $a_1^2$:
$$a_1(1+k+k^2)=21$$
$$a_1^2(1+k^2+k^4)=189$$
I want to combine these to get one equation in terms of $k$, so I will isolate the $a_1$, then substitute.
$$a_1=\frac{(1+k+k^2)}{21}$$
$$a_1^2=\frac{(1+k^2+k^4)}{189}$$
Squaring both sides in the first equation:
$$a_1^2=\frac{(1+k+k^2)}{21^2}$$
Substituting for $a_1^2$:
$$\frac{(1+k+k^2)^2}{21^2}=\frac{(1+k^2+k^4)}{189}$$
Cross multiply and combine like terms to get a polynomial in $k$:
$$0=4k^4-6k^3-2k^2-6k+4$$
$$0=2k^4-3k^3-k^2-3k+2$$
This can be factored using synthetic division and the rational roots theorem, and we find $k=2$, and can then solve for $a_1$, $a_2$, and $a_3$. I found $a_1=3$, $a_2=6$, $a_3=12$.
A: Assume that the terms are $m, mr, mr^2$.  
$(2)\div(1):$
$$\begin{align}
9&=\frac{m^2(1+r^2+r^4)}{m(1+r+r^2)}\\
&=\frac{m(1-r+r^2)(1+r+r^2)}{1+r+r^2}\\
&=m(1-r+r^2)&&\text{as }1+r+r^2\neq0\end{align}$$
and the solution for which is 
$$m=3, r=2$$
i.e. the three numbers are $3,6,12$. 
