Posted by Sidney on Monday, February 14, 2011 at 3:06pm.
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, Monday, February 14, 2011 at 3:17pm
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, Monday, February 14, 2011 at 3:23pm
Thanks, this helped a lot! :)

ERROR  math  Reiny, Monday, February 14, 2011 at 3:40pm
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, Monday, February 14, 2011 at 3:49pm
Ah, okay I see. Thanks for correcting it.
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