Suppose a_1, a_2, a_3, ldots, a_{2012} are in | vedclass

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(D) Given that each number is the average of its neighbors,we have $a_i = \frac{a_{i-1} + a_{i+1}}{2}$,which implies $2a_i = a_{i-1} + a_{i+1}$ for al

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$6036$

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(D) Given that each number is the average of its neighbors,we have $a_i = \frac{a_{i-1} + a_{i+1}}{2}$,which implies $2a_i = a_{i-1} + a_{i+1}$ for all $i$ (with indices taken modulo $2012$).This means $a_{i+1} - a_i = a_i - a_{i-1}$.Let $d_i = a_{i+1} - a_i$. Then $d_i = d_{i-1}$,which implies all differences are equal to some constant $d$.Since the numbers are arranged on a circle,$a_{2013} = a_1$,so $a_1 = a_1 + 2012d$,which implies $d = 0$.Therefore,$a_1 = a_2 = a_3 = \ldots = a_{2012} = k$ for some integer $k$.The sum of even-indexed numbers is $a_2 + a_4 + \ldots + a_{2012} = 1006k = 3018$.Thus,$k = \frac{3018}{1006} = 3$.The sum of all $2012$ numbers is $2012 	imes k = 2012 	imes 3 = 6036$.

Suppose $a_1, a_2, a_3, \ldots, a_{2012}$ are integers arranged on a circle. Each number is equal to the average of its two adjacent numbers. If the sum of all even-indexed numbers is $3018$,what is the sum of all numbers?

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