SOLUTION: The numbers n-4, n+2, 3n+1 are in geometrical progression. Find the two possible values of the common ratio.

Question 1177531: The numbers n-4, n+2, 3n+1 are in geometrical progression. Find the two possible values of the common ratio.
Answer by amarjeeth123(570) (Show Source): You can put this solution on YOUR website!
Let the first term be a.
Let the common ratio is r.
Let the three terms of the G.P.be a,ar,ar^2
Plugging in the values we get,
a=n-4.....................equation 1
ar=n+2....................equation 2
ar^2=3n+1.................equation 3
Dividing equation 2 by equation 1,r=(n+2)/(n-4)...equation 4
Substituting 1 and 4 in equation 3 we get,
(n-4)((n+2)^2/(n-4)^2)=3n+1
Simplifying we get,(n+2)^2=(3n+1)(n-4)
n^2+4n+4=3n^2-11n-4
2n^2-15n-8=0
2n^2-16n+n-8=0
2n(n-8)+1(n-8)=0
(2n+1)(n-8)=0
n=-1/2 or n=8
Substituting in equation 1 we get a=-9/2 or a=4
Substituting in equation 2 we get ar=3/2 or ar=6
r=-1/3 or r=3/2
The values of the common ratio are r=-1/3 or r=3/2
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