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