[Algebra]
Problem: The numbers $a, b, c$, in that order, form a three term arithmetic sequence (see below) and $a+b+c=60$.

The numbers $a-2, b, c+3$, in that order, form a three term geometric sequence. Determine all possible values of $a, b$ and $c$.

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7$ is an arithmetic sequence with three terms.

A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)

Present your answer in the form of coordinates (e.g. (1, 2, 3) for a=1, b=2, c=3).

Solution: Since $a, b, c$ form an arithmetic sequence, then we can write $a=b-d$ and $c=b+d$ for some real number $d$.

Since $a+b+c=60$, then $(b-d)+b+(b+d)=60$ or $3 b=60$ or $b=20$.

Therefore, we can write $a, b, c$ as $20-d, 20,20+d$.

(We could have written $a, b, c$ instead as $a, a+d, a+2 d$ and arrived at the same result.) Thus, $a-2=20-d-2=18-d$ and $c+3=20+d+3=23+d$, so we can write $a-2, b, c+3$ as $18-d, 20,23+d$.



Since these three numbers form a geometric sequence, then

$$
\begin{aligned}
\frac{20}{18-d} & =\frac{23+d}{20} \\
20^{2} & =(23+d)(18-d) \\
400 & =-d^{2}-5 d+414 \\
d^{2}+5 d-14 & =0 \\
(d+7)(d-2) & =0
\end{aligned}
$$

Therefore, $d=-7$ or $d=2$.

If $d=-7$, then $a=27, b=20$ and $c=13$.

If $d=2$, then $a=18, b=20$ and $c=22$.

(We can check that, in each case, $a-2, b, c+3$ is a geometric sequence.)
Since $a, b, c$ form an arithmetic sequence, then $c-b=b-a$ or $a+c=2 b$.

Since $a+b+c=60$, then $2 b+b=60$ or $3 b=60$ or $b=20$.

Thus, $a+c=40$, so $a=40-c$.

Therefore, we can write $a, b, c$ as $40-c, 20, c$.

Also, $a-2=40-c-2=38-c$, so we can write $a-2, b, c+3$ as $38-c, 20, c+3$.

Since these three numbers form a geometric sequence, then

$$
\begin{aligned}
\frac{20}{38-c} & =\frac{c+3}{20} \\
20^{2} & =(38-c)(c+3) \\
400 & =-c^{2}+35 c+114 \\
c^{2}-35 d+286 & =0 \\
(c-13)(c-22) & =0
\end{aligned}
$$

Therefore, $c=13$ or $c=22$.

If $c=13$, then $a=27$, so $a=27, b=20$ and $c=13$.

If $c=22$, then $a=18$, so $a=18, b=20$ and $c=22$.

(We can check that, in each case, $a-2, b, c+3$ is a geometric sequence.)

Final answer: $(27,20,13), (18,20,22)$