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Question

(A) Let the $5$ consecutive integers starting with $m$ be $m, m + 1, m + 2, m + 3, m + 4$.The average is given by: $\frac{m + (m + 1) + (m + 2) + (m +

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$\frac{2n + 5}{2}$

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(A) Let the $5$ consecutive integers starting with $m$ be $m, m + 1, m + 2, m + 3, m + 4$.The average is given by: $\frac{m + (m + 1) + (m + 2) + (m + 3) + (m + 4)}{5} = n$.Simplifying the numerator: $\frac{5m + 10}{5} = n$,which gives $m + 2 = n$.Now,we need the average of $6$ consecutive integers starting with $(m + 2)$. These integers are $(m + 2), (m + 3), (m + 4), (m + 5), (m + 6), (m + 7)$.The average is: $\frac{(m + 2) + (m + 3) + (m + 4) + (m + 5) + (m + 6) + (m + 7)}{6}$.Summing the terms: $\frac{6m + 27}{6} = m + \frac{27}{6} = m + 4.5$.Since $m = n - 2$,substitute this into the expression: $(n - 2) + 4.5 = n + 2.5$.Converting to fraction form: $n + \frac{5}{2} = \frac{2n + 5}{2}$.

The average of $5$ consecutive integers starting with $m$ is $n$. What is the average of $6$ consecutive integers starting with $(m + 2)$?

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