# 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.

Let's say the numbers are $a+d$, $a$, $a-d$, and $\dfrac{(a-d)^2}{a}$. The sum of the middle numbers is $$36=a+(a-d)=2a-d\,.$$ The sum of the other two numbers is $$37=(a+d)+\frac{(a-d)^2}{a}=(2a-d)+\frac{d^2}{a}=36+\frac{d^2}{a}\,.$$ That is, $a=d^2$. Hence, your numbers are $$d^2+d\,,\,\,d^2,d^2-d\,,\text{ and }d^2-2d+1\,.$$ It should be easy now.

You have $(2d-9)(d+4)=2d^2-d-36=(2a-d)-36=0$.

• Thank you so much! – Viktoras Bražas Sep 17 '18 at 18:43

Denoting the numbers $a,b,c,d$, we have the following system of equations:

$$\begin{cases}a+c=2b, \\bd=c^2, \\a+d=37, \\b+c=36.\end{cases}$$

We eliminate $a,b,d$ with $$b=36-c,\\a=2b-c=72-3c,\\d=37-a=3c-35$$

and finally,

$$(36-c)(3c-35)=c^2.$$

The solutions of this quadratic equation are

$$c=20\lor c=\frac{63}4$$ and the other numbers follow.

Some hints:

Call the second and third numbers $x$ and $y$. Find the first and the fourth numbers in terms of $x$ and $y$. Then write up the equations implied by the text, and solve them.

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.

$a-n + r(ra) = 37\\ a-n + r(a+n) = 37\\ a-(36-2a) + r(a+36-2a) = 37\\ 3a-36 + r(36-a) = 37\\ 3a-36 + (\frac {36}{a} - 1)(36-a) = 37\\ 3a - 36 + \frac {36^2}{a} - 36 - 36 + a = 37\\ 4a + \frac{1296}{a} - 145 = 0$

$4a^2 -145a + 1269 = 0\\ (a-16)(4a-81) = 0$
$a = 16, a = \frac {81}{4}$
$12, 16, 20, 25$
If our sequence could include rationals... $\frac {99}{4}, \frac {81}{4}, \frac {63}{4}, \frac {49}{4}$