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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 + ar^{2} = a + (a+8r) + (a+10r)
ar + ar^{2} - 18r= 2a
i also thought about doing a system of equations but don't really know where i'm going really
Thanks for your help
Arithmetic:
1st = a
9th = a + 8d
11th= a + 10d
Geometric:
1st = a
2nd = ar
3rd = ar^{2}
By equating these:
1. a=a (fairly useless!!!)
2. ar= a + 8d
a(r-1)=8d
(a/8)(r-1) = d
3. ar^{2}= a +10d
a(r^{2}-1)= 10d
(a/10)(r^{2}-1) = d
----Equate 2 and 3-----
(a/10)(r^{2}-1) = (a/8)(r-1)
8(r^{2}-1) = 10(r-1)
8(r+1)(r-1) = 10(r-1)
r+1=(10/8)
r= 2/8 = 1/4 = .25
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!