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 b-a = 9-b

from the first...

b^2 = 9a

a = b^2/9

from the second...

a = 2b-9

so b^2/9 = 2b-9

b^2 = 18b - 81

b^2 - 18b + 81 = 0

(b-9)(b-9) = 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

b-a = 9-b ---> a = 2b-9

then (2b-9)^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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