Respuesta :
Answer:
[tex]\dfrac{1}{6}$ and $\dfrac{8}{3}$ or $ \dfrac{8}{3}$ and $ \dfrac{1}{6}$ respectively.[/tex]
Step-by-step explanation:
The numbers x, [tex]\dfrac{17}{12}[/tex] and y are the first three terms of an arithmetic progression.
For three consecutive terms, a, b and c of an arithmetic sequence:
The arithmetic mean, [tex]b=\dfrac{a+c}{2}[/tex]
Therefore:
[tex]\dfrac{17}{12}=\dfrac{x+y}{2}\\12(x+y)=17*2\\12(x+y)=34\\x+y=\frac{34}{12}\\\\x= \dfrac{34}{12}-y[/tex]
The numbers x, [tex]\dfrac23[/tex] and y are three consecutive terms of a geometric progression.
For three consecutive terms, a, b and c of a geometric sequence:
The geometric mean, [tex]b=\sqrt{ac}[/tex]
Therefore:
[tex]\dfrac23=\sqrt{xy}\\xy=(\frac23)^2\\xy=\dfrac49[/tex]
Substituting [tex]x= \dfrac{34}{12}-y[/tex] derived above, we have:
[tex](\frac{34}{12}-y)y=\frac49\\\\\frac{34}{12}y-y^2=\frac49\\\\y^2-\frac{34}{12}y+\frac49=0[/tex]
This is a quadratic equation which we can then solve for y.
Using a calculator, we obtain:
[tex]y=\dfrac{1}{6}$ or $y= \dfrac{8}{3}[/tex]
[tex]When\: y=\dfrac{1}{6}, x= \dfrac{34}{12}-\dfrac{1}{6}=\dfrac{8}{3}\\\\When\: y=\dfrac{8}{3}, x= \dfrac{34}{12}-\dfrac{8}{3}=\dfrac{1}{6}[/tex]
Therefore, the values of x and y are:
[tex]\dfrac{1}{6}$ and $\dfrac{8}{3}$ or $ \dfrac{8}{3}$ and $ \dfrac{1}{6}$ respectively.[/tex]
