How to divide and set an increment within a total? I have total=90. I want to divide this number between 5 employees. 
For each employee from the second, I want to add an increment of 30% but the sum must remain 90.
Variables:
employees=5;
total=90;
increment=30%

E.g.

 Employee     Partial
   1st          10
   2nd          13
   3rd          16.9
   4th          21.7 
   5th          28.4

Is there a formula to achieve this? 
 A: If you pay the first employee an amount $P$ 
Then the total paid to the five employees will be the sum of the geometric series ...
$$T=P(1+1.3+1.3^2+ 1.3^3+1.3^4)  $$
a formula does exist for this type of sum
$$T = P \bigg(\frac{1-1.3^5}{1-1.3}\bigg)
$$
so if your total is \$90 , then the first employee should be paid 
$$P = 90\bigg(\frac{1-1.3}{1-1.3^5}\bigg) $$
Which works out to be close to \$9.95
A: So you want the first employee to get some starting amount $x$.
Then you want the second employee to get $x + 30\%$ - 30% more than the previous person. Equivalently, this is $1.3x$, obviously.
The third employee gets 30% more than the previous guy - again, $1.3$ times that amount. So they get $1.3 \times 1.3x = (1.3)^2 x$.
You can see how this goes: employee $n$ gets $(1.3)^{n-1}x$.
In that light, we could just add everything up, set it equal to $90$, combine like terms, and find $x$:
$$x + (1.3)x + (1.3)^{2}x + (1.3)^{3}x + (1.3)^{4}x = 90$$
Adding like terms and all isn't difficult here, but there's a more general way to do this for, say, REALLY long series. You see, what we have on the left of the equals sign is known as a geometric series: each term is the previous, times some number $r$:
$$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x$$
This is what we call a "finite" geometric series, because it ends after ever-so-many terms. (Some geometric series go on infinitely and they have some special nuances to deal with. You're only dealing with a finite one so I'll skip over those details.) It can be shown that the sum of a finite geometric series can be given by
$$x + rx + r^2 x+ r^3 x+ r^4 x+ r^5 x+ r^6 x +... + r^n x = \left( \frac{r^{n+1}-1}{r-1} \right) x$$
(In your case, we take $r = 1.3$.) Solving for $x$ becomes significantly easier this way.
