Understanding Consecutive Odd Integers and Averages

This problem involves finding the average of a set of consecutive odd integers. Consecutive odd integers follow each other in order, with a difference of 2 between any two successive terms. For example, 1, 3, 5, 7 are consecutive odd integers.

Finding the 5 Consecutive Odd Integers

We are given that the average of 5 consecutive odd integers in increasing order is 11.

Let the 5 consecutive odd integers be represented by variables. Since they are consecutive odd integers in increasing order, if the first integer is \(n\), the next four will be \(n+2\), \(n+4\), \(n+6\), and \(n+8\).

The average of these 5 integers is their sum divided by the count (which is 5).

Sum of the integers \( = n + (n+2) + (n+4) + (n+6) + (n+8) \)

Sum \( = 5n + (2+4+6+8) \)

Sum \( = 5n + 20 \)

The average is given as 11. So, we can set up the equation:

\(\frac{5n + 20}{5} = 11\)

To solve for \(n\), first multiply both sides by 5:

\(5n + 20 = 11 \times 5\)

\(5n + 20 = 55\)

Subtract 20 from both sides:

\(5n = 55 - 20\)

\(5n = 35\)

Divide both sides by 5:

\(n = \frac{35}{5}\)

\(n = 7\)

So, the first odd integer is 7.

The 5 consecutive odd integers are:

• First integer: \(n = 7\)
• Second integer: \(n+2 = 7+2 = 9\)
• Third integer: \(n+4 = 7+4 = 11\)
• Fourth integer: \(n+6 = 7+6 = 13\)
• Fifth integer: \(n+8 = 7+8 = 15\)

The 5 consecutive odd integers are 7, 9, 11, 13, and 15.

We can quickly check the average: \(\frac{7+9+11+13+15}{5} = \frac{55}{5} = 11\). This matches the information given in the question.

Calculating the Average of the Last 3 Integers

The question asks for the average of the last 3 of these integers. The last 3 integers are 11, 13, and 15.

To find their average, we sum these three integers and divide by 3.

Sum of the last 3 integers \( = 11 + 13 + 15 \)

Sum \( = 39 \)

Average of the last 3 integers \( = \frac{\text{Sum of last 3 integers}}{\text{Count of integers}} \)

Average \( = \frac{39}{3} \)

Average \( = 13 \)

Alternative Approach: Using Properties of Arithmetic Progressions

Consecutive odd integers form an arithmetic progression (AP) with a common difference of 2. For an AP with an odd number of terms, the average is equal to the middle term.

Since there are 5 terms, the middle term is the 3rd term.

Given the average of the 5 integers is 11, the 3rd integer is 11.

The 5 consecutive odd integers in increasing order are:

_, _, 11, _, _

Since the common difference is 2, the integer before 11 is \(11-2=9\), and the integer before 9 is \(9-2=7\).

The integer after 11 is \(11+2=13\), and the integer after 13 is \(13+2=15\).

The 5 consecutive odd integers are 7, 9, 11, 13, 15.

The last 3 integers are 11, 13, 15. These also form an AP. For these 3 terms, the average is the middle term, which is 13.

This confirms the result obtained using the algebraic method.

Summary of Integers and Averages

The average of the last 3 consecutive odd integers is 13.

Revision Table: Key Concepts

Additional Information: Properties of Averages

• The average is a measure of central tendency.
• If each number in a set is increased or decreased by a constant value, the average also increases or decreases by the same constant value.
• If each number in a set is multiplied or divided by a constant value, the average is also multiplied or divided by the same constant value.
• In a set of numbers that form an arithmetic progression, the average is equal to the average of the first and last term: \(\frac{\text{First Term} + \text{Last Term}}{2}\). For the set {7, 9, 11, 13, 15}, average is \(\frac{7+15}{2} = \frac{22}{2} = 11\). For {11, 13, 15}, average is \(\frac{11+15}{2} = \frac{26}{2} = 13\).