Question : The average of 5 consecutive integers starting with 'm' is n. What is the average of 6 consecutive integers starting with (m + 2)?
Option 1: $\frac{2n+5}{2}$
Option 2: $(n+2)$
Option 3: $(n+3)$
Option 4: $\frac{2n+9}{2}$
Correct Answer: $\frac{2n+5}{2}$
Solution :
Average of 5 consecutive integers starting with 'm' = n
Let m, m + 1, m + 2, m + 3, m + 4 be the 5 consecutive integers.
Average of 5 consecutive integers = middle term
$n$ = m + 2---------------------------------(i)
Then 6 consecutive integers starting with (m + 2)
⇒ m + 2, m + 3, m + 4, m + 5, m + 6, m + 7
From equation (i),
⇒ $n, n+1, n+2, n+3, n+4, n+5$
Average of 6 consecutive integers = average of two middle terms
= $\frac{(n+2)+(n+3)}{2}$
= $\frac{(2n+5)}{2}$
The average of 6 consecutive integers starting with (m + 2) is $\frac{(2n+5)}{2}$.
Hence, the correct answer is $\frac{(2n+5)}{2}$.
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