# AVG of INT

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

Correct result:

x =  30.5

#### Solution:

$n = 52 - 9 + 1 = 44 \ \\ s_n = \dfrac{n}{2}(a_1+a_n) = 22 \cdot (9 + 52) = 1342 \ \\ x=\dfrac{ s_n }{ n } = \dfrac{ 1342 } { 44} = \dfrac{ 9 + 52}{2} = 30.5$

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