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

anonymous · Answer

you guys go to the same school?

anonymous · Answer

just solved this with @killua_vongoladecimo

anonymous · Answer

This looks boring, a system of equations no doubt.

anonymous · Answer

just solved it five minutes ago
one way is to solve $$12x+y=x^2\12x+2y=\frac{45}{2}$$

anonymous · Answer

@satellite73 no.......i know where he got this right @Eads737 ?

anonymous · Answer

The answer is 139....ask the solution to @satellite73

anonymous · Answer

The first 3 numbers are a/r,a and ar (say)Then the fourth no. is ar+(ar-a).=2ar-a So,a/r=12 and 2ar-a=45/2 So,a=12r Therefore,212rr-12r=45/2 Solving this we get r=5/4 or -3/4 But if r= -3/4 then 2ar-a is negative So r=5/4 This yields the solution a/b=135/4 (this a is the a in the problem not the second term.) So a+b=135+4=139

anonymous · Answer

Hmm?