To solve the problem of finding the probability that three randomly drawn tickets from a set of 21 consecutively numbered tickets are in arithmetic progression (A.P.), we can follow these steps:
### Step 1: Calculate the total number of ways to choose 3 tickets from 21.
The total number of ways to choose 3 tickets from 21 is given by the combination formula:
\[
\text{Total ways} = \binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
Where \( n = 21 \) and \( r = 3 \):
\[
\text{Total ways} = \binom{21}{3} = \frac{21!}{3!(21-3)!} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1} = 1330
\]
### Step 2: Determine the number of favorable outcomes where the tickets are in A.P.
For three numbers \( a, b, c \) to be in A.P., they can be expressed as:
\[
b - a = c - b \quad \Rightarrow \quad 2b = a + c
\]
This implies that \( b \) must be the average of \( a \) and \( c \).
#### Case 1: Common difference \( d = 1 \)
The sets of numbers in A.P. with a common difference of 1 are:
- (1, 2, 3)
- (2, 3, 4)
- ...
- (19, 20, 21)
This gives us 19 valid combinations.
#### Case 2: Common difference \( d = 2 \)
The sets of numbers in A.P. with a common difference of 2 are:
- (1, 3, 5)
- (2, 4, 6)
- ...
- (17, 19, 21)
This gives us 17 valid combinations.
#### Case 3: Common difference \( d = 3 \)
The sets of numbers in A.P. with a common difference of 3 are:
- (1, 4, 7)
- (2, 5, 8)
- ...
- (15, 18, 21)
This gives us 15 valid combinations.
Continuing this process, we can find the number of valid combinations for common differences up to \( d = 10 \).
#### Summary of valid combinations:
- \( d = 1 \): 19 combinations
- \( d = 2 \): 17 combinations
- \( d = 3 \): 15 combinations
- \( d = 4 \): 13 combinations
- \( d = 5 \): 11 combinations
- \( d = 6 \): 9 combinations
- \( d = 7 \): 7 combinations
- \( d = 8 \): 5 combinations
- \( d = 9 \): 3 combinations
- \( d = 10 \): 1 combination
Adding these gives us the total number of favorable outcomes:
\[
\text{Total favorable outcomes} = 19 + 17 + 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1 = 100
\]
### Step 3: Calculate the probability.
The probability \( P \) that the three drawn tickets are in A.P. is given by the ratio of the number of favorable outcomes to the total outcomes:
\[
P = \frac{\text{Total favorable outcomes}}{\text{Total ways}} = \frac{100}{1330}
\]
### Step 4: Simplify the probability.
To simplify \( \frac{100}{1330} \):
\[
P = \frac{10}{133}
\]
### Final Answer:
The probability that the numbers on the three drawn tickets are in A.P. is:
\[
\boxed{\frac{10}{133}}
\]
