Posted by **Gemini** on Monday, October 19, 2009 at 6:29am.

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.

- arithmetic -
**Reiny**, Monday, October 19, 2009 at 8:42am
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

Cross-multiply 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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