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.
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.
Thanks, this helped a lot! :)
Ah, okay I see. Thanks for correcting it.
Well, I must say, that's quite the mathematical puzzler! Let's see if I can come up with an answer for you.
If we have a geometric progression between 3 and 9, it means that the ratio between consecutive terms is constant. And if we have an arithmetic progression between the last three terms, it means that the difference between consecutive terms is constant.
So, let's call the two missing numbers a and b. The geometric progression would be 3, a, b, 9, and the arithmetic progression would be a, b, 9.
Given this information, we can set up two equations:
b/a = 9/b (from the geometric progression)
9 - b = b - a (from the arithmetic progression)
Now, let's solve for a and b!
From the first equation, we can cross-multiply to get b^2 = 9a.
Substituting this into the second equation, we get:
9 - b = b - (b^2/9)
Simplifying further, we have:
9 = 2b - (b^2/9)
Rearranging and cross-multiplying, we get:
b^2 - 18b + 81 = 0
This equation does not yield real roots, which means there are no positive numbers that satisfy both the geometric and arithmetic progressions.
Well, I tried my best to find a solution, but it seems like this joke turned out to be a bit of a flop! Maybe math and humor just don't mix well in this case. If you have any other questions or need a good chuckle, feel free to ask!
To find the two positive numbers that can be inserted between 3 and 9 such that the first three are in geometric progression and the last three are in arithmetic progression, we can use the formulas for geometric and arithmetic progressions.
Let's start by finding the common ratio (r) of the geometric progression. We know that the first term (a) is 3 and the third term (a * r^2) is 9. So we have:
a = 3
a * r^2 = 9
Dividing the second equation by the first equation, we get:
r^2 = 9/3
r^2 = 3
r = √3 or r = -√3
Since we are dealing with positive numbers, we choose r = √3.
Now let's find the common difference (d) of the arithmetic progression. We know that the third term (9) equals the average of the first term (3) and the fourth term (d + 9). So we have:
(3 + (d + 9)) / 2 = 9
Simplifying, we get:
d + 12 = 18
d = 6
Now that we have both the common ratio (r = √3) for the geometric progression and the common difference (d = 6) for the arithmetic progression, we can find the second term (a * r) and the fifth term (9 + d) of both progressions.
For the geometric progression:
Second term (a * r) = 3 * √3
For the arithmetic progression:
Fifth term (9 + d) = 9 + 6
Sum of the two numbers = (a * r) + (9 + d)
Sum of the two numbers = (3 * √3) + (9 + 6)
Now we can calculate the sum:
Sum of the two numbers = (3 * √3) + (9 + 6)
Sum of the two numbers = (3 * √3) + 15
Therefore, the sum of the two numbers that can be inserted between 3 and 9 is (3 * √3) + 15.
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