Question

Three numbers form a geometric progression. If 4 is subtracted from the third term, then the three numbers will form an arithmetic progression. If, after this, 1 is subtracted from the second and third terms of the progression, then it will again result in a geometric progression. Find these three numbers.

Answer 1

In the first G.P. the numbers are
a, ar, ar^2
To form the A.P., the differences are constant, so

ar-a = ar^2-4 - ar

Then, forming a new G.P. with constant ratio, we have

(ar-1)/a = (ar^2-4-1)/(ar-1)

Solving for a and r, we have
a=1, r=3
or
a = 1/9, r=7

The 1st G.P. is 1, 3, 9
The A.P. is 1, 3, 5
The 2nd G.P. is 1, 2, 4

or

GP1: 1/9, 7/9, 49/9
AP: 1/9, 7/9, 13/9
GP2: 1/9, -2/9, 4/9

Answer 2

let the three numbers be a , ar, and ar^2

after 4 is subtracted from the third, we have
a , ar , and ar^2 - 4
they are now in an AP, that is
ar - a = ar^2 - 4 - ar
2ar - a - ar^2 = -4
a(2r - 1 - r^2) = -4
a = 4/(r^2 - 2r + 1) = 4/(r-1)^2

now we subtract 1 from the 2nd and 3rd, so we have
a, ar-1, and ar^2 - 5
now they form a GP again, that is ...

(ar-1)/a = (ar^2 - 5)/(ar-1)
a^2 r^2 - 2ar + 1 = a^2r^2 - 5a
1 = 2ar - 5a
1 = a(2r - 5)
1 = (4/(r-1)^2) (2r-5)
r^2 - 2r + 1 = 8r - 20
r^2 - 10r + 21 = 0
(r-7)(r-3) = 0
r = 7 or r = 3 , then
a = 1/9 or a = 1

the three original number are
1/9, 7/9, 49/9
or
1 , 3, 9

check: I will check the easy one

original GP ---> 1, 3, 9
after subtracting 4 from the last
1 , 3, 5 , which is indeed an AP

second change:
1, 2, 4, which is now a GP again
looking good!

Answer 3

1/9, 7/9, 49/9
and
1, 3, 9