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 45/2 . The sum of the two middle numbers can be written as a /b where a and b are coprime positive integers. Find a+b .

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 45/2   . The sum of the two middle numbers can be written as a /b    where a  and b  are coprime positive integers. Find a+b .

agent0smith · Answer

Maybe you can find the middle two numbers by making simultaneous equations. Use r for the common ratio of the geometric progression, and d for the common difference for the arithmetic sequence.

Start with the geometric progression... First term is 12, second term is 12r, third is 12r^2, fourth is 12r^2 + d (since only the first three are geometric) and is equal to 45/2

$$\Large 12r^2 + d = \frac{ 45 }{ 2 }$$Now use the arithmetic sequence. The second term still has to be 12r, then the third is 12r + d, fourth is 12r + 2d, and is equal to 45/2
$$\Large 12r + 2d = \frac{ 45 }{ 2 }$$
Now you can solve these for d and r, then find the middle two terms.

anonymous · Answer

again?

anonymous · Answer

twice last night, this question is like a virus