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

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- October 2nd 2006, 08:35 AMRimasGeometric progression
if the geometric progression is 216 and their sum is 19 find the largest of the three numbers

- October 2nd 2006, 11:18 AMtopsquark
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