# 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:

$\frac{b}{a}=\frac{c}{b}$

$\frac{b}{a}=\frac{4b-a}{b}$

$\frac{b}{a}=4-\frac{a}{b}$

$\frac{b}{a}+\frac{a}{b}=4$

• Definitions are your friends. Write down the two equations that corresond to each condition (one geometric sequence, one arithmetic sequence). You will most likely see how to solve this with that initial effort. Mar 9, 2017 at 23:11
• Welcome to Math.SE! With questions likes these (resembling homework), we first expect some effort from the asker before we help out. Mar 9, 2017 at 23:12
• $a, b=ar, c=ar^2$ and $a, 2b=a+d, c=a+2d$ ... & then I tried ??? Come on Jimmy. Mar 9, 2017 at 23:16
• You can absolutely use latex. And, please add your work to the original post, rather than as a comment. Mar 9, 2017 at 23:22
• Ok original post updated and latex seems to work fine. Mar 9, 2017 at 23:48

$2b-a = d\\ c - 2b = d\\ c - a = 2d\\ c+a = 4b$
$\frac {b}{a} = r\\ \frac {c}{b} = r^2\\ b = ar\\ c = ar^2$
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 $r.$