Posted by Sidney on .
There are two positive numbers that can be inserted between 3 and 9 such that the first three are in geometric progression while the last three are in arithmetic progression. Find the sum of those two numbers.

math 
Reiny,
let the sequence be
3 a b 9
then b/a = 9/b and ba = 9b
from the first...
b^2 = 9a
a = b^2/9
from the second...
a = 2b9
so b^2/9 = 2b9
b^2 = 18b  81
b^2  18b + 81 = 0
(b9)(b9) = 0
b = 9
then a = 9
sum of these two numbers is 18 
math 
Sidney,
Thanks, this helped a lot! :)

ERROR  math 
Reiny,
My solution is wrong, I misread the question.
should be
b/a = a/3  a^2 = 3b
ba = 9b > a = 2b9
then (2b9)^2 = 3b
4b^2  36b + 81 = 3b
4b^2 39b + 81 = 0
b = (39 ± √225)/8
= 27/4 or 3
if b=3, then a = 3, sum of those two is 6
if b = 27/4, then a = 9/2 or 4.5 , their sum is
however.... 3 3 3 9 does not satisfy the original condition, while
3 , 9/2 , 27/4 , 9 does work
so the 2 numbers inserted are 9/2 and 27/4
or 4.5 and 6.75
their sum is 11.25
sorry about the previous post. 
math 
Sidney,
Ah, okay I see. Thanks for correcting it.