Understanding the Average of Consecutive Positive Integers

The question asks us to determine how the average of eleven consecutive positive integers changes when the last two integers are removed. Let's break this down step by step.

We are given that the average of eleven consecutive positive integers is \(d\).

Representing Consecutive Integers

Let the first positive integer be \(n\). Since the integers are consecutive, they follow each other in order, increasing by 1 each time. The eleven consecutive positive integers can be represented as:

• \(n\)
• \(n+1\)
• \(n+2\)
• ...
• \(n+10\)

So, the list of eleven integers is \(n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8, n+9, n+10\).

Calculating the Initial Average

The average of a set of numbers is the sum of the numbers divided by the count of the numbers.

Sum of the eleven integers: We can use the formula for the sum of an arithmetic progression, which is \(\text{Sum} = \frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term})\).

Sum \( = \frac{11}{2} \times (n + (n+10)) = \frac{11}{2} \times (2n+10) = 11(n+5) \).

The initial average is \(d\). So, \(d = \frac{\text{Sum}}{\text{Number of terms}} = \frac{11(n+5)}{11} = n+5\).

Alternatively, for an odd number of consecutive integers, the average is simply the middle term. With 11 terms, the middle term is the \(\frac{11+1}{2} = 6\)th term, which is \(n+5\). So, \(d = n+5\).

The initial average is \(n+5\).

Excluding the Last Two Numbers

The last two numbers in the sequence \(n, n+1, \dots, n+10\) are \(n+9\) and \(n+10\). When these are excluded, we are left with the first nine numbers:

\(n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8\).

There are now 9 numbers remaining.

Calculating the New Average

We need to find the average of these remaining 9 numbers.

Sum of the nine integers: Using the arithmetic progression formula: \(\text{Sum} = \frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term})\).

Sum \( = \frac{9}{2} \times (n + (n+8)) = \frac{9}{2} \times (2n+8) = 9(n+4) \).

The new average is \( = \frac{\text{Sum}}{\text{Number of terms}} = \frac{9(n+4)}{9} = n+4 \).

Alternatively, for 9 consecutive integers, the average is the middle term. The middle term is the \(\frac{9+1}{2} = 5\)th term, which is \(n+4\). The new average is \(n+4\).

Determining the Change in Average

The initial average was \(d = n+5\).

The new average is \(n+4\).

Change in average = New average - Initial average

Change \( = (n+4) - (n+5) = n+4-n-5 = -1 \).

A change of \(-1\) means the average decreases by 1.

Summary of Change in Average

The average will decrease by 1.

Revision Table: Average of Consecutive Integers

Additional Information: Properties of Averages

When dealing with arithmetic progressions (like consecutive integers), removing terms from the ends affects the average predictably:

• If you remove terms symmetrically from both ends, the average remains the same.
• If you remove terms only from the higher end (as in this case), the average will decrease.
• If you remove terms only from the lower end, the average will increase.
• The amount of change depends on how many terms are removed and the common difference (which is 1 for consecutive integers).

In this specific problem, removing the two largest numbers from a set of eleven consecutive integers resulted in the average decreasing by exactly 1.