Question

The first, third and ninth terms of an arithmetic sequence form the terms of a geometric sequence. Find the common ratio of the geometric sequences.

a, a+2d, a+8d
a+2d/a = a+8d/a+2d
?

Answer 1

great so far:

(a+2d)/a = (a+8d)/(a+2d)
(a+2d)^2 = a(a+8d)

a^2 + 4ad + 4d^2 = a^2 + 8ad
4d^2 = 4ad
4d = 4a
a = d

r = (a+2d)/a
= (d + 2d)/d
= 3d/d
= 3

makes sense to me:
looks like we can start with any value for a

e.g let a = 5
then d = 5
and for the AS
t1 = 5
t3 = 5+10 = 15
t9 = 5 + 8(5) = 45
and 5, 15, 45 are in a GS, with r = 3

Answer 2

Correct