Sum of 3 numbers in GP is 39 and second number exceeds the first number by 6 . Find the numbers

let the 3 GP numbers be

a , ar, and ar^2

a+ar+ar^2 = 39
a(1+r+r^2) = 39
a = 39/(1+r+r^2)

ar - a = 6
a(r-1) = 6
a = 6/(r-1)

then 39/(1+r+r^2) = 6/(r-1)
6 + 6r + 6r^2 = 39r - 39
6r^2 - 33r + 45 = 0
2r^2 - 11r + 15 = 0
(r - 3)(2r - 5) = 0
r = 3 or r = 5/2

if r = 3 , a = 6/2 = 3
the numbers are: 3, 9, 27

if r = 5/2, a = 6/(5/2 - 1) = 4
the numbers are : 4 , 10, 25

thank u so much

To find the three numbers, we will set up a system of equations based on the given information.

Let's assume that the first number is "a", the common ratio is "r", and the second number exceeds the first number by 6. Therefore, the second number is "a + 6" and the third number is "a + 6r".

According to the given information, the sum of the three numbers is 39. Therefore, we can write the equation:

a + (a + 6) + (a + 6r) = 39

Simplifying the equation, we get:

3a + 6 + 6r = 39

Dividing both sides by 3, we have:

a + 2 + 2r = 13

Now, we need to use the fact that the second number exceeds the first number by 6, which gives us the equation:

a + 6 = (a + 6r)

Simplifying this equation, we get:

6 = 6r

Dividing both sides by 6, we find:

r = 1

Now, we can substitute the value of r back into the second equation to find the value of a:

a + 6 = (a + 6 * 1)
a + 6 = (a + 6)
a = 0

Therefore, the first number is 0.

By substituting the values of a and r into the first equation, we can find the second and third numbers:

0 + (0 + 6) + (0 + 6 * 1) = 39
0 + 6 + 6 = 39
12 = 39

Since 12 does not equal 39, there is no solution that satisfies all the given conditions.