Answer
a) $x=7,y=12$
b) $x=\sqrt[3]{68},y=\sqrt[3]{578}$
Work Step by Step
We are given the sequence:
$$2,x,y,17,\dots$$
a) In order for the sequence to be arithmetic we must have constant difference between consecutive terms:
$$x-2=y-x=17-y.$$
We obtain the system of equations:
$$\begin{cases}
x-2=y-x\\
y-x=17-y.
\end{cases}$$
Rewrite the system:
$$\begin{cases}
2x-y=2\\
-x+2y=17.
\end{cases}$$
Solve the system:
$$\begin{cases}
y=2x-2\\
-x+2(2x-2)=17.
\end{cases}$$
$$\begin{align*}
-x+4x-4&=17\\
3x&=21\\
x&=7\\
y&=2(7)-2=12.
\end{align*}$$
We got:
$$x=7,y=12.$$
b) In order for the sequence to be geometric we must have constant ratio between consecutive terms:
$$\dfrac{x}{2}=\dfrac{y}{x}=\dfrac{17}{y},x\not=0,y\not=0.$$
We obtain the system of equations:
$$\begin{cases}
x^2=2y\\
y^2=17x.
\end{cases}$$
Solve the system:
$$\begin{cases}
y=\dfrac{x^2}{2}\\
\left(\dfrac{x^2}{2}\right)^2=17x.
\end{cases}$$
$$\begin{align*}
\dfrac{x^4}{4}-17x&=0\\
x^4-68x&=0\\
x(x^3-68)&=0\\
x_1=0&\text{ or }x_2=\sqrt[3]{68}.
\end{align*}$$
As $x\not=0$, the only solution is:
$$\begin{align*}
x&=\sqrt[3]{68}\\
y&=\dfrac{(\sqrt[3]{68})^2}{2}=\sqrt[3]{578}.
\end{align*}$$
We got:
$$x=\sqrt[3]{68},y=\sqrt[3]{578}$$
