There are four numbers such that average of first two numbers is 1 more than the first number, average of first three numbers is 2 more than average of first two numbers, and average of first four numbers is 3 more than average of first three numbers. Then, the difference between the largest and the smallest numbers, is

We need to find what "Average of first two numbers is 1 more than the first number" means:

$\tfrac{p+q}{2} = p + 1$

$p + q = 2p + 2$

$q = p + 2$

The second number is always 2 more than the first number.

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For "Average of first three numbers is 2 more than average of first two numbers":

We know the average of first two numbers is $p + 1$ from the previous condition.

So: $\tfrac{p+q+r}{3} = (p + 1) + 2 = p + 3$

$p + q + r = 3p + 9$

Since we know $q = p + 2$, substituting:

$p + (p + 2) + r = 3p + 9$

$2p + 2 + r = 3p + 9$

$r = 3p + 9 - 2p - 2 = p + 7$

The third number is 7 more than the first number.

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For "Average of first four numbers is 3 more than average of first three numbers":

We know the average of first three numbers is $p + 3$ from the previous condition.

So: $\tfrac{p+q+r+s}{4} = (p + 3) + 3 = p + 6$

$p + q + r + s = 4p + 24$

Substituting our known values $q = p + 2$ and $r = p + 7$:

$p + (p + 2) + (p + 7) + s = 4p + 24$

$3p + 9 + s = 4p + 24$

$s = 4p + 24 - 3p - 9 = p + 15$

The fourth number is 15 more than the first number.

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Our four numbers are: $p$, $p + 2$, $p + 7$, $p + 15$

Notice that regardless of what $p$ is, the pattern remains the same:

Smallest number: $p$

Largest number: $p + 15$

Difference between largest and smallest = $(p + 15) - p = 15$

The differences between consecutive numbers are 2, 5, and 8, creating a specific pattern that always results in a total spread of 15.

Therefore, the difference between the largest and smallest numbers is 15.