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

The numbers are
12, 12+d, 12+2d, 452
12, 12+d, (12+d)r, (12+d)r^2=452

(12+d)r^2 = 452
12+2d = (12+d)r

so,

r = (12+2d)/(12+d)

(12+d)((12+2d)/(12+d))^2 = 452
(12+2d)^2 = 452(12+d)
d = 112.71
r = 1.09377

The numbers are

12, 124.71, 237.42, 452