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

Accepted Answer

15. Let the four numbers be $x_1, x_2, x_3, x_4$. 1) $\frac{x_1 + x_2}{2} = x_1 + 1 \implies x_2 = x_1 + 2$. Average $A_2 = x_1 + 1$. 2) $\frac{x_1 + x_2 + x_3}{3} = A_2 + 2 = x_1 + 3 \implies 2x_1 + 2 + x_3 = 3x_1 + 9 \implies x_3 = x_1 + 7$. Average $A_3 = x_1 + 3$. 3) $\frac{x_1 + x_2 + x_3 + x_4}{4} = A_3 + 3 = x_1 + 6 \implies 3x_1 + 9 + x_4 = 4x_1 + 24 \implies x_4 = x_1 + 15$. Difference between largest ($x_4$) and smallest ($x_1$) is $15$.

Averages of Consecutive Sets

Explanation

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