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  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 ...
 plz sequence sir steve reiny bob damon i need u  The three real,distint and non...
 math  The first, the third and the seventh terms of an increasing arithmetic ...
 math C2 sequences and series  The eight,fourth and second terms of an ...
 algebra  three numbers are in harmonic progression. If the third number were ...
 math  Three numbers in an arithmetic sequence sum to six. If you add 1 to the ...
 math  I would need help with example: The three numbers are consecutive terms ...
 Arithmetic  The first, second and third terms of a geometric progression are 2k...
More Related Questions