Three numbers form a geometric progression. If the second term is increased by 2, then the progression will become arithmetic and if, after this, the last term is increased by 9, then the progression will again become geometric. Find these three numbers.

Answer 1

Let the 3 numbers in GP be
a , ar, and ar^2

after first change:
a , ar+2, ar^2 are now in AP
that is,
ar+2 - a = ar^2 - (ar+2)
ar^2 - 2ar + a -4 = 0 **

after 2nd change:
a, ar+2 , ar^2 + 9 , are now in GP again,
(ar+2)/a = (ar^2 + 9)/(ar+2)
(ar+2)^2 = a(ar^2 + 9)
a^2 r^2 + 4ar + 4 = a^2 r^2 + 9a
4 = 9a - 4ar
4 = a(9-4r)
a = 4/(9-4r) ***

sub *** into **
(4/(9-4r))(r^2) - 2(4/(9-4r))(r) + 4/(9-r) = 4
times 9-4r
4r^2 - 8r + 4 = 4(9-4r)
4r^2 - 8r + 4 = 36 - 16r
4r^2 + 8r - 32 = 0
r^2 + 2r - 8 = 0
(r+4)(r-2) = 0
r = -4 or r = 2

if r = 2, a = 4/1 = 4
and the initial terms are: 4, 8, 16
check: let's add 2 to the 2nd: 4 10 16, which is AP
let's add 9 to the third: 4, 10, 25, which is GP
that works

I will let you try to see if r = -4 works

Answer 2

In the first change, how did it become ar^2 - 2ar + a -4 = 0, how was it -4? Shouldn't it be
ar+2 - a = ar^2 - (ar+2)
ar^2 - ar - ar - a + 2 - 2 =0
ar^2 - 2ar - a = 0?