To find the third term of the infinite geometric progression (GP), we need to determine the common ratio (r) and the first term (a).

Let's denote the first term as a and the common ratio as r.

Given that the sum of the infinite GP is 32, we can use the formula for the sum of an infinite GP:

S = a / (1 - r)

Substituting the given values, we have:

32 = a / (1 - r)

We are also given that the sum of the first two terms is 24. The sum of the first two terms of a GP can be calculated using the formula:

S2 = a + ar

Substituting the given values, we have:

24 = a + ar

From these two equations, we can solve for a and r simultaneously.

Solving for a in the second equation:

24 - ar = a

a = 24 / (1 - r)

Substituting this value of a into the first equation:

32 = (24 / (1 - r)) / (1 - r)

32 = 24 / (1 - r)^2

Multiplying both sides by (1 - r)^2:

(1 - r)^2 * 32 = 24

Expanding and simplifying:

32 - 64r + 32r^2 = 24

32r^2 - 64r + 8 = 0

Dividing through by 8:

4r^2 - 8r + 1 = 0

Using the quadratic formula to solve for r:

r = (-b ± √(b^2 - 4ac)) / (2a)

r = (8 ± √(64 - 16)) / (8)

r = (8 ± √48) / 8

Simplifying:

r = (8 ± 4√3) / 8

r = 1 ± √3/2

Since the common ratio r is numerically less than 1, we take the negative root:

r = 1 - √3/2

Now, substitute this value of r back into the equation for a:

a = 24 / (1 - r)

a = 24 / (1 - (1 - √3/2))

a = 24 / (√3/2)

a = 48 / √3

To find the third term, we use the formula:

T3 = a * r^2

T3 = (48 / √3) * ((1 - √3/2)^2)

T3 = (48 / √3) * (1 - 2√3/2 + 3/4)

T3 = (48 / √3) * (1 - √3 + 3/4)

T3 = (48 / √3) * (7/4 - √3)

Simplifying:

T3 = (48 * 7 - 48√3) / (4√3)

T3 = (336 - 48√3) / (4√3)

T3 = (84 - 12√3) / √3

Multiplying the numerator and denominator by √3 to rationalize the denominator:

T3 = (84√3 -