Post a New Question

arithmetic

posted by .

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 -

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question