Answer:
107
Step-by-step explanation:
Hello, Let's note n the odd number that we are looking for.
If n is odd, n+1 is even but n+2 is odd again and so on and so forth.
So, the 16 consecutive odd numbers are
n, n+2, n+4, ... n+2k, ,,,, ,n+30
It means that the average is:
[tex]\displaystyle \dfrac{1}{16}(n+(n+2)+...+(n+2k)+...+(n+2*15))\\\\=\dfrac{1}{16}\sum_{k=0}^{k=15} {(n+2k)}\\\\=\dfrac{1}{16}\left(n * 16 + 2 \sum_{k=1}^{k=15} {k} \right)\\\\=\dfrac{1}{16}\left(n * 16 + 2 (\dfrac{15*16}{2}) \right)\\[/tex]
And it must be equal to 122, so we can write.
[tex]\dfrac{1}{16}\left(n * 16 + 2 (\dfrac{15*16}{2}) \right)=122\\\\n+15=122\\\\n = 122-15=107[/tex]
Thank you.
