Question

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

Answer 1

Correct Answer: 15

Let the four numbers be $a, b, c, d$.Nbsp;

Step 1. The average of the first two numbers, $\frac{a+b}{2}$, is 1 more than $a$. This gives us:
$\frac{a+b}{2} = a+1 \implies a+b = 2a+2 \implies b = a+2$.

Step 2. The average of the first three numbers, $\frac{a+b+c}{3}$, is 2 more than the average of the first two ($\frac{a+b}{2}$). So:
$\frac{a+b+c}{3} = \frac{a+b}{2} + 2$. Since $\frac{a+b}{2} = a+1$, we have:
$\frac{a+b+c}{3} = (a+1) + 2 = a+3$.
Multiplying by 3 gives: $a+b+c = 3(a+3) = 3a+9$.
Substituting $b=a+2$: $a+(a+2)+c = 3a+9 \implies 2a+2+c = 3a+9 \implies c = a+7$.

Step 3. The average of the first four numbers, $\frac{a+b+c+d}{4}$, is 3 more than the average of the first three numbers ($\frac{a+b+c}{3}$). So:
$\frac{a+b+c+d}{4} = \frac{a+b+c}{3} + 3$. Since $\frac{a+b+c}{3} = a+3$, we have:
$\frac{a+b+c+d}{4} = (a+3) + 3 = a+6$.
Multiplying by 4 gives: $a+b+c+d = 4(a+6) = 4a+24$.
Substituting $b=a+2$ and $c=a+7$: $a+(a+2)+(a+7)+d = 4a+24 \implies 3a+9+d = 4a+24 \implies d = a+15$.

The four numbers are $a, a+2, a+7, a+15$. The largest number is $a+15$ and the smallest is $a$.
Thus, the difference between the largest and smallest is:
$(a+15) - a = 15$.