Finding the Largest of 5 Consecutive Natural Numbers

The problem asks us to find the largest number in a sequence of 5 consecutive natural numbers whose average is 43.

Understanding Consecutive Natural Numbers

Consecutive natural numbers are integers that follow each other in order, increasing by 1 each time. For example, 1, 2, 3, 4, 5 are 5 consecutive natural numbers. If we let the first number be \(n\), the next numbers will be \(n+1\), \(n+2\), \(n+3\), and \(n+4\).

Understanding Average

The average of a set of numbers is calculated by summing all the numbers and then dividing the sum by the count of numbers. In this case, we have 5 numbers.

Average = \(\frac{\text{Sum of numbers}}{\text{Count of numbers}}\)

Setting up the Equation

Let the 5 consecutive natural numbers be:

• First number: \(n\)
• Second number: \(n+1\)
• Third number: \(n+2\)
• Fourth number: \(n+3\)
• Fifth number: \(n+4\)

The sum of these 5 numbers is:

\(n + (n+1) + (n+2) + (n+3) + (n+4)\)

Combining like terms, the sum is:

\(5n + (0+1+2+3+4)\)

\(5n + 10\)

The average is given as 43. Using the average formula:

\(\text{Average} = \frac{5n + 10}{5}\)

We can simplify this expression:

\(\text{Average} = \frac{5n}{5} + \frac{10}{5}\)

\(\text{Average} = n + 2\)

We are given that the average is 43, so we can set up the equation:

\(n + 2 = 43\)

Solving for the First Number

To find the value of \(n\), we subtract 2 from both sides of the equation:

\(n = 43 - 2\)

\(n = 41\)

So, the first number in the sequence is 41.

Finding the Consecutive Numbers

Now that we know the first number (\(n=41\)), we can find the other numbers in the sequence:

• First number: \(n = 41\)
• Second number: \(n+1 = 41+1 = 42\)
• Third number: \(n+2 = 41+2 = 43\)
• Fourth number: \(n+3 = 41+3 = 44\)
• Fifth number: \(n+4 = 41+4 = 45\)

The 5 consecutive natural numbers are 41, 42, 43, 44, and 45.

Identifying the Largest Number

From the list of numbers (41, 42, 43, 44, 45), the largest number is 45.

Verification

Let's verify if the average of these numbers is indeed 43.

Sum = \(41 + 42 + 43 + 44 + 45\)

Sum = \(215\)

Average = \(\frac{215}{5}\)

Average = \(43\)

The calculated average matches the given average, confirming our numbers are correct.

The largest of these 5 consecutive natural numbers is 45.

Revision Table: Average of Consecutive Numbers

Additional Information: Properties of Consecutive Numbers

• For any set of an odd number of consecutive integers, the average is always equal to the middle number. In this case, the middle number of 41, 42, 43, 44, 45 is 43, which is the given average.
• For a set of an even number of consecutive integers, the average is the average of the two middle numbers. For example, the average of 1, 2, 3, 4 is (2+3)/2 = 2.5.
• The sum of \(k\) consecutive integers starting from \(n\) is \(kn + \frac{k(k-1)}{2}\). For 5 numbers starting from \(n=41\), \(k=5\): \(5 \times 41 + \frac{5(5-1)}{2} = 205 + \frac{5 \times 4}{2} = 205 + 10 = 215\).