Question

if the average of 14 consecutive integers is 20.5, what is the average of the first seven integers?

Answer 1

Let \(n\) be an integer, then
\[\large\frac{\displaystyle\sum_{k=0}^{13}(n+k)}{14}=20.5\]
You should be able to solve for \(n\) at this point using some known summation formulas:
\[\large\sum_{k=1}^r1=r\\
\large\sum_{k=1}^r k=\frac{r(r+1)}{2}\]
You have
\[\large\begin{align*}\sum_{k=0}^{13}(n+k)&=n+\sum_{k=1}^{13}(n+k)\\
&=n+n\sum_{k=1}^{13}1+\sum_{k=1}^{13}k\\
&=n+13n+\frac{13(13+1)}{2}\\
&=14n+91\end{align*}\]
so the first equation has become
\[\large\frac{14n+91}{14}=20.5~~\iff~~n=14\]
The average of the first seven integers would then be
\[\frac{14+15+16+17+18+19+20}{7}=\cdots\]