The $2$nd, the $1$st and the $3$rd term of an arithmetic progression form a geometric progression An arithmetic progression is given with a common difference $\ne0.$ The $2nd$, the $1st$ and the $3rd$ term of the ap form a geometric progression. Find the common ratio.
So we have the ap: $a_1,a_1+d,a_1+2d$ and the gp: $a_1+d,a_1,a_1+2d.$ How can we find $q?$ Thank you in advance!
I have tried to solve the problem using a system, but I am not sure which equations we can use. For example, $a_1^2=(a_1+d)(a_1+2d)$ and $\dfrac{a_1+d}{2}=a_1+(a_1+2d)$ hold, but I don't see how to use them.
Why is it important for the common difference of the AP to be $\ne0?$ Then we will have a constant sequence in which all terms are equal, right?

The three terms of a geometric series $a_0, a_1= a_0 + d, a_2 = a_0+2d$ or if it is easier $a_0= a_1-d, a_1, a_2= a_1 + d$.
The three terms of a geometric series is $b_0, b_1=b_0\cdot r, b_2=b_0\cdot r^2$ or if it is easier $b_0=\frac {b_1}r, b_1, b_2= b_1\cdot r$.
We are told that $a_1, a_1 -d, a_1 +d$ form a geometric series. so
$\frac {b_1}r = a_1$.  $b_1 = a_1-d$, $b_1r = a_1 + d$.
So Solve for $r$.
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substitute $b_1$ with $a_1 -d$ and
$\frac {a_1 -d}r = a_1$ and $(a_1 -d)r = a_1 + d$.
So $r = \frac {a_1 -d}{a_1} = 1 - \frac d{a_1}$ and $r = \frac {a_1 + d}{a_1-d}=1 + \frac {2d}{a_1-d}$
So $-\frac d{a_1} = \frac {2d}{a_1-d}$ so $-\frac 1{a_1} = \frac 2{a_1-d}$ so
$-(a_1 -d) = 2a_1$ so $d = 3a_1$.
Substituting $3a_1$ for $d$ we get
$r = \frac {a_1 -d}{a_1}  = \frac {-2a_1}{a_1} = -2$ and  $r = \frac {a_1 + d}{a_1-d}= \frac {4a_1}{-2a_1} = -2$.
So $r = -2$.
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Furthermore the three terms as an arithmetic series $a_0 = a_1-d =-2a_1; a_1 = a_1; a_2= a_1 + d = 4a_1$.
And as a geometric series they are $b_0 = a_1; b_1 =-2a_1; b_2 = 4a_1$.
