If the average of 9 consecutive positive integers is 55, then what is the largest integer?
59
Finding the Largest Consecutive Integer Given the Average
Let's solve this problem step-by-step. We are given that the average of 9 consecutive positive integers is 55. We need to find the largest of these integers.
When you have a set of consecutive integers, their average is the middle number if the total count of integers is odd. In this case, we have 9 consecutive integers, and 9 is an odd number.
So, the average, which is 55, is exactly the middle integer among the 9 integers.
Since there are 9 integers, the middle integer is the one with an equal number of integers before and after it. If we list them from 1st to 9th, the middle one is the 5th integer.
Number of integers = 9
Middle term position = \((9 + 1) / 2 = 10 / 2 = 5\)th term
So, the 5th integer in the sequence is 55.
The integers are consecutive, meaning they follow each other in order, increasing by 1 each time. The 5th integer is 55.
To find the largest integer, which is the 9th integer, we need to add the difference in positions to the middle term. There are \(9 - 5 = 4\) integers after the 5th integer.
Since the integers are consecutive, each position increases the value by 1. So, the 6th integer is \(55+1\), the 7th is \(55+2\), the 8th is \(55+3\), and the 9th (the largest) is \(55+4\).
Largest integer = 5th integer + (Number of terms after the 5th term)
Largest integer = \(55 + (9 - 5)\)
Largest integer = \(55 + 4\)
Largest integer = 59
Let's list the integers to verify:
- 1st integer: \(55 - 4 = 51\)
- 2nd integer: \(55 - 3 = 52\)
- 3rd integer: \(55 - 2 = 53\)
- 4th integer: \(55 - 1 = 54\)
- 5th integer: 55 (the average)
- 6th integer: \(55 + 1 = 56\)
- 7th integer: \(55 + 2 = 57\)
- 8th integer: \(55 + 3 = 58\)
- 9th integer: \(55 + 4 = 59\) (the largest)
The sequence of 9 consecutive positive integers is 51, 52, 53, 54, 55, 56, 57, 58, 59. The largest integer is 59.
Revision Table: Average of Consecutive Integers
| Property | Description |
|---|---|
| Average (Odd number of terms) | The average is the middle term. |
| Average (Even number of terms) | The average is the average of the two middle terms. |
| Finding terms from middle | For consecutive integers, add/subtract 1 for each step away from the middle term. |
Additional Information on Consecutive Integers
Consecutive integers are integers that follow each other in order, differing by 1. For example, 1, 2, 3 or 10, 11, 12.
Let the first integer be \(n\). The next consecutive integers would be \(n+1, n+2, n+3, \dots\).
If you have an odd number of consecutive integers, say \(2k+1\) integers, the average is the \((k+1)\)-th term.
If you have an even number of consecutive integers, say \(2k\) integers, the average falls between the \(k\)-th and \((k+1)\)-th term. The average is the sum of the \(k\)-th and \((k+1)\)-th term divided by 2.
In our problem, we had 9 integers (\(2k+1 = 9 \implies 2k=8 \implies k=4\)), so the average was the \((4+1)=5\)th term.
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