Question 11.

Let us assume the four numbers to be a, b, c and d in ascending order. 

Average of first two numbers is 1 more than the first number

$\frac{\left(a+b\right)}{2}=a+1$

$b-a=2$

$b=a+2$

Average of first three numbers is 2 more than average of first two numbers

$\frac{\left(a+b+c\right)}{3}=\frac{\left(a+b\right)}{2}+2$

$2c=a+b+12$

Substituting the value for b

$2c=a+a+2+12$

$2c=2a+14$

$c=a+7$

Average of first four numbers is 3 more than average of first three numbers.

$\frac{\left(a+b+c+d\right)}{4}=\frac{\left(a+b+c\right)}{3}+3$

$3d=a+b+c+36$

Substituting the value of b and c

$3d=a+a+2+a+7+36$

$3d=3a+45$

$d=a+15$

d is the largest and a is the smallest and we know that d=a+15

Hence the difference between the smallest and the largest values is 15.