Geometric and arithmetic sequence Let a,b and c be consecutive terms in a geometric sequence and a, 2b and c be consecutive terms in an arithmetic sequence. Determine the quotient b/a. 
$a,2b,c$ arithmetic sequence
$2b=a+d$ and $c=a+2d$
And, $a+2b+c=3(a+c)/2\leftrightarrow2a+4b+2c=3a+3c\leftrightarrow$
$4b=a+c$ 
$a,b,c$ geometric sequence
$b=ar$ and $c=ar^{2}$
And, $a+b+c=\frac{a(1-r^{3})}{1-r}$
Came to this first:
$\frac{b}{a}=\frac{c}{b}$
$\frac{b}{a}=\frac{4b-a}{b}$
$\frac{b}{a}=4-\frac{a}{b}$
$\frac{b}{a}+\frac{a}{b}=4$

$2b-a = d\\
c - 2b = d\\
c - a = 2d\\
c+a = 4b$
That is about as much as we can say right now about the arithmetic sequence.  Lets look at the geometric sequence.
$\frac {b}{a} = r\\
\frac {c}{b} = r^2\\
b = ar\\
c = ar^2$
We want to solve for $r$
Substitute: $b= ar, c=ar^2$ into $a+c = 4b$  
$ar^2+a = 4ar\\
a(r^2 - 4r + 1) = 0$
Use the quadratic formula to solve for $r.$
