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 $\displaystyle \frac {2k}{3} - 2$ but can't figure it out..)
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 $\displaystyle \frac {2k}{3} - 2$ but can't figure it out..)
Perhaps the idea of an arithmetic sequence can help us.
Let $\displaystyle n$ be the middle number so that our sequence is $\displaystyle (n-2)+n+(n+2) = k$
For 3 consecutive odd numbers the mean will be equal to the middle number. Hence
$\displaystyle n = \frac{k}{3}$ is the middle number.
Hence the smaller number is: $\displaystyle n-2 = \frac{k}{3} - 2$.
The question is asking us to find the sum of the two smaller numbers:
$\displaystyle (n-2)+n = \left(\frac{k}{3} - 2 \right) + \frac{k}{3} = \frac{2k}{3} - 2$
which is the book's answer
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 $\displaystyle n=3$ and hence $\displaystyle k = 1+3+5 = 9$
$\displaystyle (3-2)+3 = \frac{2(9)}{3} - 2$ and sure enough $\displaystyle 4=4$
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