Solution:

To prove that the 29th term of an arithmetic progression (AP) is double the 19th term when the 9th term is zero, we can use the formula for the nth term of an AP.

Formula:
The nth term of an AP is given by the formula:
an = a + (n-1)d

Where:
an = nth term
a = first term
n = position of the term
d = common difference

Given information:
9th term (a9) = 0

Step 1: Find the common difference (d)
To find the common difference, we can use the formula for the difference between two consecutive terms in an AP:

d = a(n+1) - an

Substituting the given values:
0 = a(9+1) - a9
0 = a10 - a9

Since a9 is zero, we can simplify the equation to:
0 = a10

This implies that the 10th term is also zero.

Step 2: Find the first term (a)
To find the first term, we can use the formula:
a = an - (n-1)d

Substituting the known values:
a = a9 - (9-1)d
a = 0 - 8d
a = -8d

Step 3: Finding the 19th term (a19)
Using the formula for the nth term:
a19 = a + (19-1)d
a19 = -8d + 18d
a19 = 10d

Step 4: Finding the 29th term (a29)
Using the formula for the nth term:
a29 = a + (29-1)d
a29 = -8d + 28d
a29 = 20d

Step 5: Proving the statement
We need to prove that the 29th term (a29) is double the 19th term (a19).
So, we need to show that:
a29 = 2 * a19

Substituting the values:
20d = 2 * 10d

Since the common difference (d) is non-zero, we can cancel it out from both sides of the equation:
20 = 2 * 10

This equation is true, which proves that the 29th term is indeed double the 19th term in the given arithmetic progression.