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Suppose a_{1},a_{2},a_{3}........a_{2012} are integers arrange

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?

0

1509

3018

6036

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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?

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Text solutionVerified

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

∣ a_{1} ∣ = ∣ a_{2} ∣ = ∣ a_{3} ∣ = ∣ a_{4} ∣ = ...... = ∣ a_{2012} ∣

3018 = a_{2} + a_{4} + a_{6} + a_{8} + ...... + a_{2012} 3018 = 1006 ∣ a ∣ 3 = ∣ a ∣

Total

S u m = a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + ...... + a_{2012} S_{n} = (2012) ∣ a ∣ S_{n} = 3 \times 2012 S_{n} = 6036

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Question Text	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?
Answer Type	Text solution:1
Upvotes	150