AVG of INT

What is the average of the integers from 9 throuht 52 inclusive?

Correct answer:

x =  30.5

Step-by-step explanation:

$$\begin{gathered} n=52-9+1=44 \\ {s}_n={\displaystyle \frac{n}{2}}(a_1+a_n)=22\cdot (9+52)=1342 \\ x={\displaystyle \frac{s_n}{n}}={\displaystyle \frac{1342}{44}}={\displaystyle \frac{9+52}{2}}=30.5 \end{gathered}$$

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