# Thread: geometric and arithmetic progression

1. ## geometric and arithmetic progression

hey, just need a hint if you don't mind mates!
The first 3 terms of a geometric progression are also the 1st, 9th , 11th terms of an arithmetic progression.
Given that the terms of the geometric progression are all different, find the common ratio, r.

I've started with: a + ar + ar2 = a + (a+8r) + (a+10r)
ar + ar2 - 18r= 2a

i also thought about doing a system of equations but don't really know where i'm going really

up

3. ## Re: geometric and arithmetic progression

Arithmetic:
1st = a
9th = a + 8d
11th= a + 10d

Geometric:
1st = a
2nd = ar
3rd = ar2

By equating these:
1. a=a (fairly useless!!!)

2. ar= a + 8d
a(r-1)=8d
(a/8)(r-1) = d

3. ar2= a +10d
a(r2-1)= 10d
(a/10)(r2-1) = d

----Equate 2 and 3-----

(a/10)(r2-1) = (a/8)(r-1)
8(r2-1) = 10(r-1)
8(r+1)(r-1) = 10(r-1)
r+1=(10/8)
r= 2/8 = 1/4 = .25

4. ## Re: geometric and arithmetic progression

Why do you think that the phrase "The first 3 terms of a geometric progression are also the 1st, 9th , 11th terms of an arithmetic progression" means that the sum of the first 3 terms of a geometric progression equals the sum of the 1st, 9th , 11th terms of the arithmetic progression? Also, why do you think that the ratio of the geometric progression equals the common difference of the arithmetic progression?

When you correctly express this phrase algebraically, you'll get two equations and three unknowns. Indeed, this system is satisfied by infinitely many triples (a, r, d) where d is the common difference, but still there is only one possible r.

Edit: People, this forum is not for offering detailed solutions!

5. ## Re: geometric and arithmetic progression

ohh i see thank you very much