IIT JEE , AIEEE Forum - Mathematics , Algebra

JimboJones

Basic idea for solving the program.

$\mbox{Given } a_1 , a_2, \cdots.a_n \mbox{ such that } a_1 \leq a_2 , a_2 \le a_3, \cdots, a_{n-1} \leq a_n$

$\mbox{along with the condition that } a_1 \mbox{ is the airthmetic mean of } a_2, a_3,\cdots, a_n$

$\mbox{For this condition to hold, everyone of } a_2, a_3, \cdots, a_n \mbox{ must be equal to } a_1$

$\mbox{Else the mean is more than } a_i \mbox{ , which is a contradiction}$

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Let us assume that numbers arranged in the squares are not all equal. Thus, the numbers have a minium number. Consider a square that has this number. The, number in the square is the average of the numbers in the neighbouring squares. Since the number under consideration is the minimum, all the numbers in the neighouring squares is equal to the minimum number. Repeating this process in the neigbouring squares, we get that all the numbers are equal.

QED.

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