The average of 5 consecutive integers starting with m as the first int...
Approach: Work with a numerical example
The fastest way to solve problems of this kind is to take numerical examples.
Let the 5 consecutive integers be 1, 2, 3, 4, and 5.
The average of 5 consecutive integers from 1 to 5 is 3.
Therefore, the value of m is 1 and the value of n is 3.
Because m = 1, m + 2 = 3
9 consecutive integers starting from m + 2 will be 3, 4, ... , 11
The average of positive integers from 3 to 11 is 7.
We know n = 3. So, 7 has to be (n + 4).
Choice E is the correct answer.
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The average of 5 consecutive integers starting with m as the first int...
Understanding the Problem
To find the average of 5 consecutive integers starting with m, we first define these integers:
- m, m+1, m+2, m+3, m+4
The average (n) of these integers can be calculated as:
- Average (n) = (m + (m+1) + (m+2) + (m+3) + (m+4)) / 5
This simplifies to:
- n = (5m + 10) / 5 = m + 2
Finding the Average of 9 Consecutive Integers
Now, we need to find the average of 9 consecutive integers starting with (m + 2):
- The integers are: (m + 2), (m + 3), (m + 4), (m + 5), (m + 6), (m + 7), (m + 8), (m + 9), (m + 10)
The average of these integers is:
- Average = ((m + 2) + (m + 3) + (m + 4) + (m + 5) + (m + 6) + (m + 7) + (m + 8) + (m + 9) + (m + 10)) / 9
This simplifies to:
- Average = (9m + 54) / 9 = m + 6
Relating it to n
Since we previously found that n = m + 2, we can express the average of the 9 integers in terms of n:
- Average = m + 6 = (n - 2) + 6 = n + 4
Conclusion
Thus, the average of the 9 consecutive integers starting with (m + 2) is:
- n + 4
This confirms that the correct answer is option 'E'.