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.

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?

Let's assume that the three numbers in the geometric progression are a, ar, and ar^2, where a is the first term and r is the common ratio.

According to the given information:

1. If the second term (ar) is increased by 2, the progression becomes arithmetic. This means the common difference between consecutive terms is constant. Therefore, the second term of the new arithmetic progression is (ar + 2).

2. If, after this, the last term is increased by 9, the progression becomes geometric again. This means the common ratio between consecutive terms is constant. Therefore, the last term of the new geometric progression is (ar + 2) + 9 = (ar + 11).

Since the second term in the new arithmetic progression is equal to the first term in the new geometric progression, we can set up the following equation:

ar + 2 = ar + 11

Simplifying this equation, we get:

2 = 11

This is not possible, which means there is no solution to the given problem.

To solve this problem, let's start by assuming that the three numbers form a geometric progression, with the first term being 'a' and the common ratio being 'r'. So, the three terms can be written as 'a', 'ar', and 'ar^2'.

We are given that, if the second term is increased by 2, then the progression becomes arithmetic. This means that the second term, 'ar', when increased by 2, should be equal to the average of the first and third terms.

So, we have the equation: (ar) + 2 = (a + ar^2) / 2

Now, let's simplify the equation.

Multiply both sides by 2: 2ar + 4 = a + ar^2
Rearrange the terms: ar^2 - 2ar + (4 - a) = 0

This is a quadratic equation in terms of 'r'. To solve this, we can use the quadratic formula: r = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = -2, and c = (4 - a).

Substituting these values into the quadratic formula:

r = (-(-2) ± √((-2)^2 - 4(1)(4 - a))) / (2 * 1)
= (2 ± √(4 - 4(4 - a))) / 2
= (2 ± √(4 - 16 + 4a)) / 2
= (2 ± √(-12 + 4a)) / 2
= 1 ± √(-3 + a)

Since we assumed 'r' to be a positive number, we can discard the negative solution.

Now, let's consider the condition that if the last term is increased by 9, then the progression becomes geometric again. This means that the third term, 'ar^2', when increased by 9, should be equal to the product of the first and second terms.

So, we have the equation: ar^2 + 9 = a * ar

Now, let's substitute the positive value of 'r' we found earlier:

(a * (1 + √(-3 + a)))^2 + 9 = a * a * (1 + √(-3 + a))

Expanding and simplifying the equation:

a^2 + 2a√(-3 + a) + (-3 + a) + 9 = a^2 + a√(-3 + a)
2a√(-3 + a) + 6 = a√(-3 + a)
2√(-3 + a) = √(-3 + a)
2 = 1

This is a contradiction, meaning that there is no solution to this problem.

Therefore, there are no three numbers that satisfy all the given conditions.

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