Given:
The 24th term of an arithmetic progression (AP) is twice the 10th term.

To prove:
The 72nd term of the AP is equal to four times the 15th term.

Proof:

Step 1: Understanding the arithmetic progression (AP)
An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is denoted by 'd'.

Step 2: Expressing the terms of the AP in terms of the first term (a) and the common difference (d)
The nth term of an AP can be expressed as:
Tn = a + (n - 1)d

Step 3: Expressing the given information in terms of the first term and the common difference
Let the 10th term of the AP be T10.
T10 = a + 9d

The 24th term of the AP is twice the 10th term.
T24 = 2T10
T24 = 2(a + 9d)

Step 4: Finding the relationship between the 10th term and the 24th term
Using the expressions for T10 and T24, we can equate them to find the relationship between a and d.
a + 9d = 2(a + 9d)
a + 9d = 2a + 18d
a = 9d

Step 5: Finding the relationship between the 15th term and the first term
Using the expression for a in terms of d, we can express the 15th term (T15) in terms of the first term.
T15 = a + 14d
T15 = 9d + 14d
T15 = 23d

Step 6: Expressing the 72nd term in terms of the first term and the common difference
Using the expression for Tn, we can find the value of the 72nd term (T72) in terms of a and d.
T72 = a + 71d
T72 = 9d + 71d
T72 = 80d

Step 7: Proving that the 72nd term is four times the 15th term
We need to prove that T72 = 4T15.

Substituting the expressions for T72 and T15, we have:
80d = 4(23d)
80d = 92d

Since the value of d is the common difference, 80 is equal to 92. This is not true, which means the assumption that T72 = 4T15 is incorrect.

Conclusion:
Based on the calculations, we can conclude that the 72nd term of the given arithmetic progression is not equal to four times the 15th term.