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July 31, 2014

July 31, 2014

Posted by **lana** on Wednesday, July 18, 2007 at 10:57am.

the general expression for consecutive multiples of 6 is 6N, 6(N + 1), 6(N +2), etc. find three consecutive multiples of 6 such that 4 times the first exceeds twice the third by 12.

and than Bobpursley replied:

Let N be the first, so n+1 is next, etc.

4(6(n+1))-2(6(n+3))=12

so, find n, then 6(n+1) for the first, 6(n+2) for the second, and 6(n+3) for the third.

So now i dont understand what he means by that. help me out here

Did you solve for n?

If you did, then you need three consecutive numbers divisible by six. They will be:

6(n+1) is the first number.

6(n+2) will be the second number.

You figure out what the third number will be.

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