Question

In a sequence of four positive numbers, the first three are in geometric progression and the last three are in arithmetic progression. The first number is 12 and the last number is 452. The sum of the two middle numbers can be written as ab where a and b are coprime positive integers. Find a+b.

Answer 1

@phi

Answer 2

yes, I looked at that, and I don't see how to get integers.

this sequence works
\[ 12 , 3+ \sqrt{2721}, \frac{455+ \sqrt{2721}}{2}, 452 \]

Answer 3

the last 3 are in arith progression, so
12r, 12r+d, 12r+2d

12r+2d= 452
6r+d = 226
d= 226-6r

use that in the 3rd number:
12r^2 = 12r+d
12r^2 = 12r + 226 - 6r

solve for r... you get what I posted.... but those numbers are irrational, so I don't understand the rest of the question...

a*b won't work, and if ab means concatenation, that still does not work...so I am at a loss

Answer 4

I don't know what ab means... if you can get an answer with *any* interpretation of what ab means, I would be amazed.