There are $4$ numbers. First $3$ make an arithmetic progression. Last $3$ make a geometric progression. 
There are four numbers. The first three make an arithmetic progression, and the last three make a geometric progression. The sum of first and last number is $37$. The sum of middle numbers is $36$. Find the numbers. 

So I'm trying to solve this problem for about 2 hours now. I can't use any formulas because all I get $0=0$. I don't know all the terms in English so I'll just try to upload my notes here. Eh but it doesn't let me. At first i got that $a_1=-2d$, $d$ was $= -36$ so my numbers were $72$, $36$, $0$, $-35$. The problem is that last 3 numbers doesn't make a geometric progression.

The first 3 are in arithmetic progression
$a-n, a, a+n, \cdots$
the last 3 are in geometric progression.
$\cdots, a, ra, r^2 a$
There are 4 numbers total.
That means that the 3d element must fit both the geometric and the arithmetic progression
$ra = a+n$
The middle two sum to 36
$a+ra=a(r+1) = 36\\
r = \frac{36}{a} - 1\\
36 -2a = n$
The first and last sum to 37
$a-n + r^2 a = 37$
And now we use the substitutions above to get everything in terms of a.
$a-n + r(ra) = 37\\
a-n + r(a+n) = 37\\
a-(36-2a) + r(a+36-2a) = 37\\
3a-36 + r(36-a) = 37\\
3a-36 + (\frac {36}{a} - 1)(36-a) = 37\\
3a - 36 + \frac {36^2}{a} - 36 - 36 + a = 37\\
4a + \frac{1296}{a} - 145 = 0$
Which multiplies into a quadratic.
$4a^2 -145a + 1269 = 0\\
(a-16)(4a-81) = 0$
$a = 16, a = \frac {81}{4}$
If our sequence is integers....
$12, 16, 20, 25$
If our sequence could include rationals...
$\frac {99}{4}, \frac {81}{4}, \frac {63}{4}, \frac {49}{4}$
