Posted by **John Berkhamp** on Tuesday, July 23, 2013 at 1:29pm.

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.

- Algebra/Number Theory -
**Steve**, Tuesday, July 23, 2013 at 2:00pm
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

## Answer this Question

## Related Questions

- Maths - 1..The first 2 terms of a geometric progression are the same as the ...
- math - The first, the third and the seventh terms of an increasing arithmetic ...
- math - There are two positive numbers that can be inserted between 3 and 9 such ...
- math - I would need help with example: The sum of three consecutive terms of ...
- Math (Geometric Progression) - 5 distinct positive reals form an arithmetic ...
- Math - 5 distinct positive reals form an arithmetic progression. The 1st, 2nd ...
- math - I would need help with example: The three numbers are consecutive terms ...
- math - Three numbers in an arithmetic sequence sum to six. If you add 1 to the ...
- Arithmetic - The first, second and third terms of a geometric progression are 2k...
- maths - three number form an arithmetric sequence. Their sum is 24. A)If 'a' is ...

More Related Questions