An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3, 5, 7, 9 is an arithmetic sequence with four terms.

A geometric sequence is a sequence in which each term after the first is obtained by multiplying the previous term by a constant. For example, $3,6,12,24$ is a geometric sequence with four terms.

(a) Determine a real number $w$ for which $\frac{1}{w}, \frac{1}{2}, \frac{1}{3}, \frac{1}{6}$ is an arithmetic sequence.

(b) Suppose $y, 1, z$ is a geometric sequence with $y$ and $z$ both positive. Determine all real numbers $x$ for which $\frac{1}{y+1}, x, \frac{1}{z+1}$ is an arithmetic sequence for all such $y$ and $z$.

(c) Suppose that $a, b, c, d$ is a geometric sequence and $\frac{1}{a}, \frac{1}{b}, \frac{1}{d}$ is an arithmetic sequence with each of $a, b, c$, and $d$ positive and $a \neq b$. Determine all possible values of $\frac{b}{a}$.
(a) For the sequence $\frac{1}{w}, \frac{1}{2}, \frac{1}{3}, \frac{1}{6}$ to be an arithmetic sequence, it must be the case that

$$
\frac{1}{2}-\frac{1}{w}=\frac{1}{3}-\frac{1}{2}=\frac{1}{6}-\frac{1}{3}
$$

Since $\frac{1}{3}-\frac{1}{2}=\frac{1}{6}-\frac{1}{3}=-\frac{1}{6}$, then $\frac{1}{2}-\frac{1}{w}=-\frac{1}{6}$ and so $\frac{1}{w}=\frac{1}{2}+\frac{1}{6}=\frac{2}{3}$, which gives $w=\frac{3}{2}$.

(b) The sequence $\frac{1}{y+1}, x, \frac{1}{z+1}$ is arithmetic exactly when $x-\frac{1}{y+1}=\frac{1}{z+1}-x$ or $2 x=\frac{1}{y+1}+\frac{1}{z+1}$.

Since $y, 1, z$ is a geometric sequence, then $\frac{1}{y}=\frac{z}{1}$ and so $z=\frac{1}{y}$. Since $y$ and $z$ are positive, then $y \neq-1$ and $z \neq-1$.

In this case, $\frac{1}{y+1}+\frac{1}{z+1}=\frac{1}{y+1}+\frac{1}{\frac{1}{y}+1}=\frac{1}{y+1}+\frac{y}{1+y}=\frac{y+1}{y+1}=1$.

Since $\frac{1}{y+1}+\frac{1}{z+1}=1$, then the sequence $\frac{1}{y+1}, x, \frac{1}{z+1}$ is arithmetic exactly when $2 x=1$ or $x=\frac{1}{2}$.

(c) Since $a, b, c, d$ is a geometric sequence, then $b=a r, c=a r^{2}$ and $d=a r^{3}$ for some real number $r$. Since $a \neq b$, then $a \neq 0$. (If $a=0$, then $b=0$.)

Since $a \neq b$, then $r \neq 1$. Note that $\frac{b}{a}=\frac{a r}{a}=r$ and so we want to determine all possible values of $r$.

Since $a$ and $b$ are both positive, then $r>0$.

Since $\frac{1}{a}, \frac{1}{b}, \frac{1}{d}$ is an arithmetic sequence, then

$$
\begin{aligned}
\frac{1}{b}-\frac{1}{a} & =\frac{1}{d}-\frac{1}{b} \\
\frac{1}{a r}-\frac{1}{a} & =\frac{1}{a r^{3}}-\frac{1}{a r} \\
\frac{1}{r}-1 & =\frac{1}{r^{3}}-\frac{1}{r} \quad(\text { since } a \neq 0) \\
r^{2}-r^{3} & =1-r^{2} \\
0 & =r^{3}-2 r^{2}+1 \\
0 & =(r-1)\left(r^{2}-r-1\right)
\end{aligned}
$$

Since $r \neq 1$, then $r^{2}-r-1=0$.

By the quadratic formula, $r=\frac{1 \pm \sqrt{(-1)^{2}-4(1)(-1)}}{2}=\frac{1 \pm \sqrt{5}}{2}$.

Since $a$ and $b$ are both positive, then $r>0$ and so $r=\frac{1+\sqrt{5}}{2}$.

This is the only possible value of $r$.

We can check that $r$ satisfies the conditions by verifying that when $a=1$ (for example) and $r=\frac{1+\sqrt{5}}{2}$, giving $b=\frac{1+\sqrt{5}}{2}, c=\left(\frac{1+\sqrt{5}}{2}\right)^{2}$, and $d=\left(\frac{1+\sqrt{5}}{2}\right)^{3}$, then we do indeed obtain $\frac{1}{b}-\frac{1}{a}=\frac{1}{d}-\frac{1}{b}$.