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 idexed numbers is `3018`, what is the sum of all numbers ?

Options

A. `0`
B. `9054`
C. `12072`
D. `6036`

Explanation

To solve the problem step by step, we will use the information given in the question and the properties of averages.

### Step 1: Understanding the Problem
We have a sequence of integers \( a_1, a_2, a_3, \ldots, a_{2012} \) arranged in a circle. Each number is equal to the average of its two adjacent numbers. This means:
\[ a_n = \frac{a_{n-1} + a_{n+1}}{2} \]
for all \( n \), where the indices wrap around (i.e., \( a_0 = a_{2012} \) and \( a_{2013} = a_1 \)).

### Step 2: Setting Up the Equations
From the problem, we know the sum of all even-indexed numbers:
\[ S_{even} = a_2 + a_4 + a_6 + \ldots + a_{2012} = 3018 \]
Since there are 1006 even-indexed terms (from 2 to 2012), we can express this sum as:
\[ S_{even} = \sum_{k=1}^{1006} a_{2k} = 3018 \]

### Step 3: Expressing the Even-Indexed Terms
Next, we multiply the sum of the even-indexed terms by 2:
\[ 2S_{even} = 2(a_2 + a_4 + a_6 + \ldots + a_{2012}) = 6036 \]

### Step 4: Relating Even-Indexed Terms to Odd-Indexed Terms
Each even-indexed term can be expressed in terms of its adjacent odd-indexed terms:
- For \( a_2 \):
\[ 2a_2 = a_1 + a_3 \implies a_1 + a_3 = 2a_2 \]
- For \( a_4 \):
\[ 2a_4 = a_3 + a_5 \implies a_3 + a_5 = 2a_4 \]
Continuing this way, we can express the sum of all even-indexed terms in terms of odd-indexed terms:
\[ 2S_{even} = (a_1 + a_3) + (a_3 + a_5) + (a_5 + a_7) + \ldots + (a_{2011} + a_1) \]

### Step 5: Summing Up the Terms
This can be simplified to:
\[ 2S_{even} = 2(a_1 + a_3 + a_5 + \ldots + a_{2011}) \]
Thus, we can write:
\[ S_{odd} = a_1 + a_3 + a_5 + \ldots + a_{2011} \]
So we have:
\[ 2S_{odd} = 6036 \implies S_{odd} = 3018 \]

### Step 6: Finding the Total Sum
Now, we can find the total sum of all numbers:
\[ S_{total} = S_{even} + S_{odd} = 3018 + 3018 = 6036 \]

### Final Answer
Thus, the sum of all numbers \( a_1 + a_2 + a_3 + \ldots + a_{2012} \) is:
\[ \boxed{6036} \]