1. ## Geometric progression

if the geometric progression is 216 and their sum is 19 find the largest of the three numbers

2. Originally Posted by Rimas
if the geometric progression is 216 and their sum is 19 find the largest of the three numbers
A geometric series is defined as the numbers:
a_{n+1} = r*a_n for n = 0, 1, 2, ....

where some a_0 is known.

As an explicit formula for the a_n this becomes:
a_n = c*r^n
where c is the constant c = a_0.

So we have three numbers in geometric progression and their product is 216 and their sum is 19:
[c*r^n]*[c*r^{n+1}]*[c*r^{n+2}] = 216
c*r^n + c*r^{n+1} + c*r^{n+2} = 19

We can, without loss of generality set n = 0.

[c]*[c*r]*[c*r^2] = 216
c + c*r + c*r^2 = 19

or
c^3*r^3 = 216 => c*r = 6 so we have that c = 6/r

So
c(1 + r + r^2) = 19

(6/r)(1 + r + r^2) = 19

6 + 6r + 6r^2 = 19r

6r^2 - 13r + 6 = 0

Which has solutions:
r = 3/2 and r = 2/3

Thus c = 6/r so c = 4 or c = 9 respectively.

Thus the two possible series are:
4, 6, 9 (4*6*9=216 and 4+6+9=19 check!)
9, 6, 4 (9*6*4=216 and 9+6+4=19 check!)
(Yes, these ARE two distinct series.)

Either way, the largest number in the series is 9.

-Dan