The average of 5 consecutive integers startin | vedclass

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Question

(A) The $5$ consecutive integers starting with $m$ are $m, m+1, m+2, m+3, m+4$.Their average is $\frac{m + (m+1) + (m+2) + (m+3) + (m+4)}{5} = n$.$\fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The $5$ consecutive integers starting with $m$ are $m, m+1, m+2, m+3, m+4$.Their average is $\frac{m + (m+1) + (m+2) + (m+3) + (m+4)}{5} = n$.$\frac{5m + 10}{5} = n \Rightarrow m + 2 = n \Rightarrow m = n - 2$.Now,we need the average of $6$ consecutive integers starting with $(m+2)$.These integers are $(m+2), (m+3), (m+4), (m+5), (m+6), (m+7)$.Their average is $\frac{(m+2) + (m+3) + (m+4) + (m+5) + (m+6) + (m+7)}{6} = \frac{6m + 27}{6} = m + \frac{27}{6} = m + 4.5$.Substituting $m = n - 2$ into the expression:Average $= (n - 2) + 4.5 = n + 2.5 = \frac{2n + 5}{2}$.

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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