The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

Do I need to use substitution somehow?? (The answer given is but can't figure it out..)

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- Mar 28th 2010, 12:59 PMmjoshuaconsecutive odd integers
The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

Do I need to use substitution somehow?? (The answer given is but can't figure it out..) - Mar 28th 2010, 01:06 PMmr fantastic
- Mar 28th 2010, 01:07 PMe^(i*pi)
- Mar 28th 2010, 01:14 PMmjoshua
That IS the entire question (it's taken out of my book as is) :(

- Mar 28th 2010, 01:30 PMe^(i*pi)
Perhaps the idea of an arithmetic sequence can help us.

Let be the middle number so that our sequence is

For 3 consecutive odd numbers the mean will be equal to the middle number. Hence

is the middle number.

Hence the smaller number is: .

The question is asking us to find the sum of the two smaller numbers:

which is the book's answer - Mar 28th 2010, 10:04 PMmjoshua
- Mar 29th 2010, 11:09 AMe^(i*pi)
Neither 0 nor 2 are odd integers.

If you think about an arithmetic sequence it is defined as there being a common difference between each term. For consecutive odd integers this is 2. n is 2 more than n-2 and n+2 is 2 more than n.

If we try and hence

and sure enough - Mar 29th 2010, 09:47 PMmjoshua
But instead of declaring "n" just an odd integer, shouldn't we use a sequence like 2n + 1 so that no matter what you put it for n (even or odd) you still get an odd integer? (so it's more inclusive?)

- Mar 30th 2010, 02:22 AMmr fantastic
- Mar 30th 2010, 04:49 AMbjhoppersum of odd integers
Hello Mjoshua,

I was going to reply to this post yesterday but decided that eipi had adequately answered it I passed.Today you appear confused.Here is the way I would have replied.

Three cons odd integers are n, n+2, n+4, the sum of these =k

Solution of this equation n=k-6/3. The sum of the first two is 2n+2 and thisbecomes 2/3k-2 when you insert the value of n in terms of k.

bjh - Mar 30th 2010, 10:01 AMmjoshua