The sum of three consecutive terms of an Arithmetic progression is $18$ and their product is $120$. Find the terms. I'm currently going through some questions on series and stumbled on this. 

The sum of three consecutive terms of an Arithmetic progression is $18$ and their product is $120$. Find the terms.

I attempted it like this but I think I'm wrong cause the products is giving me $210$. 

$x + (x+1) +(x+2) = 18$; 

$3x +3 = 18;$

$3x = 15;$

$x=5,$ The three consecutive numbers are 

$5, 6, 7$; 

Appreciate the assistance.
In general, the consecutive terms can differ by any number, say $d$, instead of just $1$. So, the three numbers are better represented as $x-d,x,$ and $x+d$.

Then,
$$3x=18 \Rightarrow x=6$$
and
$$x(x^2-d^2)=120$$
$$6(36-d^2)=120$$
$$36-d^2=20$$
$$d^2=16$$
$$d=\pm4$$

The numbers thus are $\{2,6,10\}$ or $\{10,6,2\}$.
Let $a_0=0$ and $a_1=1$. Then let the arithmetic progression for your sequence be

$a_n = a_(n-1) + a_(n-2)$

You will get a sequence {3,5,8} which 3*5*8=120

This is the Fibonacci sequence.