SOLUTION: Twelve consecutive integers are arranged in order from least to greatest. If the sum of the first six integers is 381, what is the sum of the last six integers?

Question 486660: Twelve consecutive integers are arranged in order from least to greatest. If the sum of the first six integers is 381, what is the sum of the last six integers?
Found 2 solutions by chessace, bucky:
Answer by chessace(471) (Show Source): You can put this solution on YOUR website!
In order, each in 2nd half is 6 more than corresponding # in 1st half,
So sum = 381 + 6 * 6 = 417.

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
The integers are consecutive. Therefore, each integer is obtained by adding 1 to the immediately preceding integer.
.
So if n+0 is the first integer, then n+1 is the second integer and the third integer is n+1+1 or n+2, and so on. The first six of the integers are:
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n + 0
n + 1
n + 2
n + 3
n + 4
n + 5
.
The sum of these 6 numbers is 6n + (0+1+2+3+4+5) which simplifies to 6n + 15.
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The problem states that the sum of the first 6 numbers is 381. So we can set up the equation:
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6n + 15 = 381
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Get rid of the 15 on the left side by subtracting 15 from both sides as follows:
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6n + 15 - 15 = 381 - 15
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On the left side the + 15 and the -15 cancel each other out and on the right side 381 - 15 = 366. So the equation simplifies to:
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6n = 366
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solve for n by dividing both sides by 6 to get:
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n = 366/6 = 61
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We now know that the first number is 61. Therefore, because they are consecutive, the first 6 numbers are 61, 62, 63, 64, 65, and 66. That means that the seventh through twelfth numbers are 67, 68, 69, 70, 71, and 72. Adding these 6 numbers:
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67 + 68 + 69 + 70 + 71 + 72
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results in the total of 417 and that's the answer to this problem.
.
Hope that this helps you to understand the problem.

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