Answer
a. $x=7, \quad y=15$
b. $x=2^{2/3}\cdot 17^{1/3},\quad y=2^{1/3}\cdot 17^{2/3}$
Work Step by Step
a.
An arithmetic sequence has the nth term
$a_{n}=a+(n-1)d$,
where a is the first term, d is the common differrence.
From $a_{4}=17$, we find d:
$17=2+3d$
$15=3d$
$5=d$
So, $x=2+5=7$
$y=7+5=12$
$x=7, \quad y=15$
b.
A geometric sequence has the nth term
$a_{n}=ar^{n-1}$
where a is the first term, d is the common ratio.
From $a_{4}=17$, we find $r$:
$ 17=2\cdot r^{3}$
$\displaystyle \frac{17}{2}=r^{3}$
$r=(\displaystyle \frac{17}{2})^{1/3}$
So,
$ x=2\cdot (\displaystyle \frac{17}{2})^{1/3}=2^{2/3}\cdot 17^{1/3},\quad $
$ y=2\cdot (\displaystyle \frac{17}{2})^{2/3}=2^{1/3}\cdot 17^{2/3}$
