Three integers

For $a$ , $b$ and $c$ to be in an arithmetic progression, $\implies b = \dfrac {a+c}2$ or $a+c = 2b$ . From $a+b+c = 3$ $\implies 3b=3$ $\implies b =1$ .

Let $r$ be the common ratio of the geometric progression. If the order of GP is $a \to b \to c$ , then:

$\begin{aligned} \frac 1r + 1 + r & = 3 \\ \frac 1r + r & = 2 \\ r^2 -2r + 1 & = 0 \\ (r-1)^2 & = 0 \\ \implies r & = 1\end{aligned}$

Then $a=b=c=1$ and they are not distinct and not acceptable.

If the order of GP is $b \to a \to c$ , then:

$\begin{aligned} 1+r+r^2 & = 3 \\ r^2 +r -2 & = 0 \\ (r+2)(r-1) & = 0 \\ \implies r & = -2 & \small \color{#3D99F6} \text{Since }r=1 \text{ is unacceptable.} \end{aligned}$

$\implies a = -2, \ b = 1, \ c=4$ $\implies abc = \boxed{-8}$

As $a,b,c$ is an arithmetic progression we have:

$a=b-r$

$c=b+r$

It's useful to define the midpoint of a series as the variable because, as in this case, it often results in the possibility of cancelling.

$a+b+c=(b-r)+b+(b+r)=3$

$3b=3$

$b=1$

We therefore have 3 numbers: $1-r$ , $1$ and $1+r$ . The only way these can be in geometric progression is if one is negative, and the ratio of the geometric progression is also negative. The obvious way for this to happen is if the progression is: $1$ , $1-r$ , $1+r$ .

Let $m$ be the ratio:

$m=1-r$

$m^2=1+r$

Squaring the first equation gives us:

$m^2=1-2r+r^2$

Equating:

$1-2r+r^2=1+r$

$r(3-r)=0$

So $r=3$ , $a=-2,b=1,c=4$ , and $\boxed{abc=-8}$