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.
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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