Question

Suppose $a_1,a_2,a_3........a_{2012}$ 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?

• A. $0$
• B. $1509$
• C. $3018$
• D. $6036$

Answer 1

The correct option is D $6036$

$S_n=a_1+a_2+a_3+a_4+a_5+......+a_{2012}$

Number of even terms $=\frac{2012}{2}=1006$
Number of odd terms $=1006$

$\left|a_1\right|=\left|a_2\right|=\left|a_3\right|=\left|a_4\right|=......=\left|a_{2012}\right|$

$3018=a_2+a_4+a_6+a_8+......+a_{2012}3018=1006\left|a\right|3=\left|a\right|$

Total $Sum=a_1+a_2+a_3+a_4+a_5+......+a_{2012}{S}_n=\left(2012\right)\left|a\right|S_n=3\times 2012S_n=6036$