Question

I would need help with example: The sum of three consecutive terms of geometric progression is 9. The first number with no change, the second number plus 12 and the third number minus 3, are the 3 consequtive terms of arithmetic progression. What are the values of original 3 terms?
I can not move with it. I would be grateful for any help.
Thank you.

Answer 1

Let the 3 terms as a GP be

a , ar, and ar^2
a + ar + ar^2 = 9
a(r^2 + r + 1) = 9 , #1

new 3 terms:
a
ar+12
ar^2 - 3


now they are in a AS
ar+12 - a = ar^2-3 - (ar+12)
ar - a + 12 = ar^2 - ar - 15
ar^2 - 2ar + a = 27
a(r^2 - 2r + 1) = 27 , #2

divide #2 by #1

(r^2 - 2r + 1)/(r^2 + r + 1 = 27/9 = 3
3r^2 + 3r + 3 = r^2 - 2r + 1
2r^2 + 5r + 2 = 0
(2r + 1)(r + 2) = 0

r = -1/2 or r = -2

if r = -2, in #1,
a(4-2+1) = 9
a = 9/3=3
The 3 original terms are 3 , -6 , 12

if r = -1/2 , in #1
a(1/4 - 1/2 + 1) = 9
a((3/4) = 9
a = 12

or the 3 original terms were 12 , -6 , 3

Answer 2

I did get first two equations but didn´t know how to continue. Thank you very much for help.