SOLUTION: the three consecutive terms in a the G.P are the second third of and sixth term of an A.P find the common ratio?

Algebra ->  Sequences-and-series -> SOLUTION: the three consecutive terms in a the G.P are the second third of and sixth term of an A.P find the common ratio?      Log On


   

Question 1083648: the three consecutive terms in a the G.P are the second third of and sixth term of an A.P find the common ratio?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let's define a couple of variables:
b= the first of the three consecutive terms in the G.P.
r= the common ratio of the G.P.
Then, the next two (consecutive) terms in the G.P are
b%2Ar= the second of the three consecutive terms in the G.P. ,
and b%2Ar%5E2= the third of the three consecutive terms in the G.P.
The problem probably meant to tell us that
b= the second term of the A.P.
b%2Ar= the third term of the A.P.
That makes b%2Ar-b the common difference of the A.P.
The problem probably meant to tell us that
b%2Ar%5E2= the sixth term of the A.P.
That would make it 4 common differences more than the second term of the A.P.:
b%2Ar%5E2=b%2B4%2A%28b%2Ar-b%29
We can simplify and solve for r the equation above
b%2Ar%5E2=b%2B4%2A%28b%2Ar-b%29
b%2Ar%5E2=b%2B4b%2Ar-4b%29
b%2Ar%5E2-4b%2Ar=-3b
b%2Ar%5E2-4b%2Ar%2B3b=0
b%28r%5E2-4r%2B3%29=0
b%28r-1%29%28r-3%29=0
We have 3 possible options:
1) b=0 and r is undefined,
probably not the answer expected.
All the terms of the A.P. and the G.P. would be zero.
2) r=1 , also probably not the answer expected.
All the terms of the A.P. and the G.P. would be b, whatever that value is.
3) highlight%28r=3%29 is probably the expected answer.