arithmetic
posted by Gemini .
I have a Q that im trying to solve:
it says: the 1st , 2nd, and 3rd term of a geo. sequence is the same as the 1st, 7th and 9th terms of an arith. sequence.: I have to find the common ratio.
So i know: a+8d = a(r^2) (1)
a+6d = ar (2)
but i don't know how to solve this simulataneously! if I divide 1 with 2 i get r = (a+8d) / (a+6d) don't knwo how I should solve it.. iknow the answer is (1/3) in the book. please help.

ok, let's call your first result of
r = (a+8d) / (a+6d) #1 equation
You could also subtract the two equations to get
ar^2  ar = 2d
r^2  r = 2d/a let's call that #2
Here is where it gets messy...
sub #1 into #2
(a+8d)^2/(a+6d)^2  (a+8d)/(a+6d) = 2d/a
finding a common denominator of (a+6d)^2 and simpifying the left side, I got
(2ad + 16d^2)/(a+6d)^2 = 2d/a
divide both sides by 2d
(a+8d)/(a+6d)^2 = 1/a
Crossmultiply and simplify to get
a^2 + 12ad + 3d^2 = a^2 + 8ad
36d^2 + 4ad = 0
9d^2 + ad = 0
d(9d + a) = 0
d = 0 (the arithmetic sequence would not change) OR
d = 9a
back into #1
r = (a+8d)/(a+6d)
= (9d + 8d)/(9d + 6d)
= d/3d
= 1/3
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