The average of 5 consecutive integers starting with "n" is m...
5 consecutive integers → n, n+1, n+ 2, n+3, n+4
As the series in AP, the middle term is average
∴ m = n+2
6 consecutive integers starting with (n+2)
→ (n+2), (n+3), (n+4), (n+5), (n+6), (n+7)
Average of 6 consecutive integers
= Average of 1st and last terms
The average of 5 consecutive integers starting with "n" is m...
Understanding the Problem
To solve the problem, we need to find the average of 6 consecutive integers starting from (n + 2), given that the average of 5 consecutive integers starting from "n" is "m".
Step 1: Average of 5 Consecutive Integers
The 5 consecutive integers starting from "n" are:
- n, n+1, n+2, n+3, n+4
To find their average:
- Average = (n + (n + 1) + (n + 2) + (n + 3) + (n + 4)) / 5
- Average = (5n + 10) / 5
- Average = n + 2
Since we know this average equals "m":
- m = n + 2
- Therefore, n = m - 2
Step 2: Average of 6 Consecutive Integers
Now, consider the 6 consecutive integers starting from (n + 2):
- (n + 2), (n + 3), (n + 4), (n + 5), (n + 6), (n + 7)
To find their average:
- Average = ((n + 2) + (n + 3) + (n + 4) + (n + 5) + (n + 6) + (n + 7)) / 6
- Average = (6n + 27) / 6
- Average = n + 4.5
Step 3: Substitute for n
Using n = m - 2 from our earlier calculation:
- Average = (m - 2) + 4.5
- Average = m + 2.5
To express this in terms of "m":
- Average = (2m + 5) / 2
Conclusion
The average of the 6 consecutive integers starting from (n + 2) is (2m + 5) / 2, which corresponds to option A.