Solution:

Let's assume the first term of the geometric progression (GP) is 'a' and the common ratio is 'r'.

Sum of an Infinite GP:
The sum of an infinite GP can be calculated using the formula:

S = a / (1 - r)

Given that the sum of the infinite GP is 15, we can write the equation as:

15 = a / (1 - r) ----(1)

Sum of Squares of the GP:
The sum of the squares of the terms in a GP can be calculated using the formula:

S = a^2 / (1 - r^2)

Given that the sum of the squares of the GP is 45, we can write the equation as:

45 = a^2 / (1 - r^2) ----(2)

Using Equation (1) and Equation (2) to solve for 'a' and 'r':
To find the values of 'a' and 'r', we need to solve equations (1) and (2) simultaneously.

From Equation (1), we can express 'a' in terms of 'r' as:

a = 15(1 - r) ----(3)

Substituting the value of 'a' from Equation (3) into Equation (2), we get:

45 = (15(1 - r))^2 / (1 - r^2)

Simplifying the equation:

45(1 - r^2) = 15^2(1 - r)^2

Dividing both sides by 15^2:

3(1 - r^2) = (1 - r)^2

Expanding and simplifying:

3 - 3r^2 = 1 - 2r + r^2

Rearranging the terms:

4r^2 - 2r - 2 = 0

Dividing both sides by 2:

2r^2 - r - 1 = 0

Using the quadratic formula to solve for 'r':
The quadratic formula is given by:

r = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 2, b = -1, and c = -1. Substituting these values into the quadratic formula, we get:

r = (-(-1) ± √((-1)^2 - 4(2)(-1))) / (2(2))
r = (1 ± √(1 + 8)) / 4
r = (1 ± √9) / 4

Simplifying:

r = (1 ± 3) / 4

Therefore, we have two possible values for 'r':

r₁ = 4/4 = 1
r₂ = -2/4 = -1/2

Finding the values of 'a' and 'r' for each case:
Case 1: r = 1
Substituting the value of 'r' into Equation (1), we get:

15 = a / (1 - 1)
15 = a / 0

This equation is undefined since we cannot divide by 0. Therefore, this case is not valid.

Case 2: r