The 1st, 2nd and 3rd terms of an exponential sequence are the 2nd, 3rd and 6th terms of a linear sequence. Find the common ratio.
Naturally the bot is WRONG again
For the AS,
term2 = a+d
term3 = a+2d
term6 = a+5d
for the GS:
(a+2d)/(a+d) = ((a+5d)/(a+2d)
a^2 + 6ad + 5d^2 = a^2 + 4ad + 4d^2
2ad + d^2 = 0
d(2a + d) = 0
d = 0 , not a very interesting sequence, or
d = -2a
So there is no unique solution, we can pick any a, that will give us d, etc
e.g. let a = 3, then d = -6 , then the AS is 3, -3, -9, -15, -21, -27 , and sure
enough the 2nd, 3rd and 6th terms, which are -3,-9,-27 form a GS with
a common ratio of 3
let a = -5, then d = 10 , then the AS is -5, 5, 15, 25, 35, 45 , and sure
enough the 2nd, 3rd and 6th terms, which are 5, 15, 45 form a GS with
a common ratio of 3
in general:
if a = a , d = -2a, then the AS is a, -a, -3a, -5a, -7a, -9a, .. and sure
enough the 2nd, 3rd and 6th terms, which are -a, -3a, -9a form a GS with
a common ratio of 3
so the common ratio for the GS is 3