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They want me to find the 3 members of a decreasing geometric progression. And I am given that let them be $a,b,c$ $a+b+c=14$ and that $a^2+b^2+c^2=84$

I tried to express everything with the first so $a+a.q+a.q^2=14$

$a(1+q+q^2)=14$

$a^2(1+q^2+q^4)=84$ (I tried to solve these to equations but couldn't. I can clearly see that the members are probably $8,4,2$ and $q=0,5$ but I can't show it.

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  • $\begingroup$ Better use $a/q+a+aq$ which makes it much easier $\endgroup$ – user35508 Mar 19 '17 at 17:17
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$$\begin{align} a(1+r+r^2)&=14&&\cdots (1)\\ a^2(1+r^2+r^4)&=84\\\Rightarrow a^2(1+r+r^2)(1-r+r^2)&=84&&\cdots (2)\\\\ (2)\div (1):\qquad\qquad\qquad a(1-r+r^2)&=6&&\cdots(3)\\\\ ((1)- (3))\div2:\qquad\qquad\qquad\qquad \;\; ar&=4&&\cdots (4)\\ ((1)+ (3))\div2:\qquad\qquad\qquad a(1+r^2)&=10&&\cdots (5)\\\\ (5)\div (4):\qquad\qquad\qquad \frac {1+r^2}{r}&=\frac 52\\ 2r^2-5r+2&=0\\ (2r-1)(r-2)&=0\\ \because r<0 \text{ (decreasing GP) }\qquad\therefore r&=\frac 12\\ a&=8 \end{align}$$ Hence GP is $$\color{red}{8,4,2}$$

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An interesting trick comes from noticing that $\frac{1+q^2+q^4}{1+q+q^2} = 1-q+q^2$, hence $$ a(1+q+q^2) = 14,\qquad a^2(1+q^2+q^4)=84 $$ lead to $a(1-q+q^2)=\frac{84}{14}=6$, then to $2aq=14-6=8$, hence $aq=4$ and $q$ is a root of $$ 1+q+q^2 = 14\cdot\frac{q}{4} $$ so the only chances are $q=2$ and $q=\frac{1}{2}$ and it is straightforward to finish.

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