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# mathcounts

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Suppose x,y, and z form a geometric sequence. If you know that and x + y + z = 18 , find the value of $x^2+y^2+z^2=612$.

Jul 23, 2019

#1
+23872
+2

Suppose $$x$$, $$y$$, and $$z$$ form a geometric sequence.
If you know that and $$x + y + z = 18$$,
find the value of
$$x^2+y^2+z^2=612$$.

I assume:

Geometric sequence:
$$x = a \\ y = ar \\ z = ar^2$$

$$\begin{array}{|lrcll|} \hline & \mathbf{x+y+z} &=& \mathbf{18} \\ &a+ar+ar^2 &=& 18 \\ &a(1+r+r^2) &=& 18 \quad |\quad 1+r+r^2 = \dfrac{1-r^3}{1-r} \qquad {\displaystyle |r|<1} \\\\ (1) &\mathbf{a\left(\dfrac{1-r^3}{1-r}\right)} &=& \mathbf{18} \qquad \text{or} \quad \mathbf{a} = \mathbf{18\left(\dfrac{1-r}{1-r^3}\right)} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \mathbf{x^2+y^2+z^2} &=& \mathbf{612} \\ &a^2+a^2r^2+a^2r^4 &=& 612 \\ &a^2(1+r^2+r^4) &=& 612 \quad &|\quad 1+r^2+r^4 = \dfrac{1-r^6}{1-r^2} \qquad {\displaystyle |r|<1} \\ (2)&\mathbf{a^2\left(\dfrac{1-r^6}{1-r^2}\right)} &=& \mathbf{612} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline \dfrac{(2)}{(1)}: & \dfrac{ a^2\left(\dfrac{1-r^6}{1-r^2}\right) } { a\left(\dfrac{1-r^3}{1-r}\right)} &=& \dfrac{612}{18} \\\\ & \dfrac{ a\left(\dfrac{1-r^6}{1-r^2}\right) } { \left(\dfrac{1-r^3}{1-r}\right)} &=& 34 \\\\ & a \dfrac{(1-r^6)(1-r)}{(1-r^3)(1-r^2)} &=& 34 \quad |\quad 1-r^6 = (1-r^3)(1+r^3) \\\\ & a \dfrac{(1-r^3)(1+r^3)(1-r)}{(1-r^3)(1-r^2)} &=& 34 \quad |\quad 1-r^2 = (1-r)(1+r) \\\\ & a \dfrac{(1-r^3)(1+r^3)(1-r)}{(1-r^3)(1-r)(1+r)} &=& 34 \\\\ (3)&\mathbf{a\left(\dfrac{1+r^3}{1+r}\right)} &=& \mathbf{34} \qquad \text{or} \quad \mathbf{a} = \mathbf{34\left(\dfrac{1+r}{1+r^3}\right)} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1)=(3): & a=18\left(\dfrac{1-r}{1-r^3}\right) &=& 34\left(\dfrac{1+r}{1+r^3}\right) \\\\ & 18\left(\dfrac{1-r}{1-r^3}\right) &=& 34\left(\dfrac{1+r}{1+r^3}\right) \\\\ & 18(1-r)(1+r^3) &=& 34(1+r)(1-r^3) \quad & | \quad : 2 \\ & 9(1-r)(1+r^3) &=& 17(1+r)(1-r^3) \\ & 9(1-r+r^3-r^4) &=& 17(1+r-r^3-r^4) \\ & 9-9r+9r^3-9r^4 &=& 17+17r-17r^3-17r^4 \\ & 8r^4+26r^3-26r-8 &=& 0 \quad & | \quad : 2 \\ & \mathbf{4r^4+13r^3-13r-4} &=& \mathbf{0} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline \text{Solutions: }\\ r &=& -1 & \text{no solution} \quad |r|< 1! \\ r &=& 1 & \text{no solution} \quad |r|< 1! \\ \mathbf{ r } &=& \mathbf{ \dfrac{-13+\sqrt{105}}{8} }\ \checkmark & (r=-0.34413115426) \\ \mathbf{ r } &=& \mathbf{ \dfrac{-13-\sqrt{105}}{8} }\ \checkmark & (r=-2.90586884574) \\ \hline \end{array}$$

First solution: $$\mathbf{ r =\dfrac{-13+\sqrt{105}}{8} }$$

$$\begin{array}{|rcll|} \hline \mathbf{a(1+r+r^2)} &=& \mathbf{18} \\\\ a &=& \dfrac{18}{1+r+r^2} \\\\ a &=& \dfrac{18}{ 1+ \dfrac{\sqrt{105}-13}{8}+ \left( \dfrac{\sqrt{105}-13}{8} \right)^2 } \\ \ldots \\ a &=& \dfrac{64}{13-\sqrt{105}} * \left( \dfrac{13+\sqrt{105}}{13+\sqrt{105}} \right) \\\\ \mathbf{a} &=& \mathbf{13+\sqrt{105}} & (a=23.2469507660) \\\\ x &=& a \\ \mathbf{x} &=& \mathbf{13+\sqrt{105}} \\\\ y &=& ar \\ y &=& \left(13+\sqrt{105}\right) \left(\dfrac{-13+\sqrt{105}}{8} \right) \\ y &=& -\left(13+\sqrt{105}\right) \left(\dfrac{13-\sqrt{105}}{8} \right) \\ y &=& \dfrac{-64}{8} \\ \mathbf{y} &=& \mathbf{-8} \\\\ z = ar^2 &=& yr \\ z &=& -8\left( \dfrac{-13+\sqrt{105}}{8} \right) \\ \mathbf{z} &=& \mathbf{13-\sqrt{105}} & (z=2.75304923404) \\ \\ \mathbf{x^2+y^2+z^2} &=& \mathbf{612} \\ \left(13+\sqrt{105}\right)^2 + (-8)^2 + \left(13-\sqrt{105}\right)^2 &=& 612 \ \checkmark \\ \hline \end{array}$$

Second solution: $$\mathbf{ r =\dfrac{-13-\sqrt{105}}{8} }$$

$$\begin{array}{|rcll|} \hline \mathbf{a(1+r+r^2)} &=& \mathbf{18} \\\\ a &=& \dfrac{18}{1+r+r^2} \\\\ a &=& \dfrac{18}{1+\dfrac{-13-\sqrt{105}}{8}+\left(\dfrac{-13-\sqrt{105}}{8}\right)^2} \\ \ldots \\ a &=& \dfrac{64}{13+\sqrt{105}} * \left( \dfrac{13-\sqrt{105}}{13-\sqrt{105}} \right) \\\\ \mathbf{a} &=& \mathbf{13-\sqrt{105}}& (a=2.75304923404) \\\\ x &=& a \\ \mathbf{x} &=& \mathbf{13-\sqrt{105}} \\\\ y &=& ar \\ y &=& \left(13-\sqrt{105}\right) \left(\dfrac{-13-\sqrt{105}}{8} \right) \\ y &=& -\left(13-\sqrt{105}\right) \left(\dfrac{13+\sqrt{105}}{8} \right) \\ y &=& \dfrac{-64}{8} \\ \mathbf{y} &=& \mathbf{-8} \\\\ z = ar^2 &=& yr \\ z &=& -8\left( \dfrac{-13-\sqrt{105}}{8} \right) \\ \mathbf{z} &=& \mathbf{13+\sqrt{105}} & (z=23.2469507660) \\ \\ \mathbf{x^2+y^2+z^2} &=& \mathbf{612} \\ \left(13-\sqrt{105}\right)^2 + (-8)^2 + \left(13+\sqrt{105}\right)^2 &=& 612 \ \checkmark \\ \hline \end{array}$$

Jul 24, 2019