The average of 50 consecutive natural numbers is x. What will be the new average when the next four numbers are also included?
x + 2
Calculating the New Average of Consecutive Natural Numbers
This problem asks us to find the new average of a sequence of consecutive natural numbers after adding more numbers to the sequence. We are given the initial average and the number of terms added.
Understanding Averages of Consecutive Numbers
For a sequence of consecutive natural numbers (or any arithmetic progression), the average is always the middle term. If the number of terms is even, the average is the average of the two middle terms. Alternatively, the average is also the average of the first and last term.
Initial Scenario: 50 Consecutive Numbers
We start with 50 consecutive natural numbers. Let the first number be \(n\). The sequence is \(n, n+1, n+2, \ldots, n+49\).
The number of terms is 50.
The average of these 50 consecutive numbers is given as \(x\).
Using the property that the average is the mean of the first and last terms:
Initial Average \(x = \frac{n + (n+49)}{2}\)
\(x = \frac{2n + 49}{2}\)
\(x = n + \frac{49}{2}\)
\(x = n + 24.5\)
Adding the Next Four Numbers
The next four consecutive natural numbers after \(n+49\) are \(n+50, n+51, n+52, n+53\).
These four numbers are added to the original sequence.
The new sequence of consecutive natural numbers is \(n, n+1, n+2, \ldots, n+49, n+50, n+51, n+52, n+53\).
New Scenario: 54 Consecutive Numbers
The total number of terms in the new sequence is \(50 + 4 = 54\).
The first term is \(n\).
The last term is \(n+53\).
We need to find the new average of these 54 consecutive numbers.
New Average \( = \frac{\text{Sum of 54 terms}}{54}\)
Using the property that the average is the mean of the first and last terms for a consecutive sequence:
New Average \( = \frac{n + (n+53)}{2}\)
New Average \( = \frac{2n + 53}{2}\)
New Average \( = n + \frac{53}{2}\)
New Average \( = n + 26.5\)
Finding the Change in Average
The initial average was \(x = n + 24.5\).
The new average is \(n + 26.5\).
Let's find the difference between the new average and the initial average \(x\).
New Average \( - x = (n + 26.5) - (n + 24.5)\)
New Average \( - x = n + 26.5 - n - 24.5\)
New Average \( - x = 26.5 - 24.5\)
New Average \( - x = 2\)
So, the New Average \( = x + 2\).
Alternative Perspective: Adding Consecutive Terms
When you add \(k\) consecutive terms to the end of an existing sequence of \(N\) consecutive terms, the average increases by exactly \(k/2\). In this problem:
- Initial number of terms, \(N = 50\).
- Number of consecutive terms added, \(k = 4\).
- The increase in average is \(k/2 = 4/2 = 2\).
Therefore, the new average will be the original average plus 2.
New Average \( = x + 2\).
Both methods lead to the same result. The new average when the next four consecutive natural numbers are included is \(x + 2\).
Revision Table: Consecutive Numbers and Averages
| Concept | Description | Average Calculation |
|---|---|---|
| Consecutive Natural Numbers | Numbers that follow each other in order, differing by 1 (e.g., 5, 6, 7) | |
| Average of Odd Number of Consecutive Terms | The middle term of the sequence. | \( \frac{\text{First Term} + \text{Last Term}}{2} \) |
| Average of Even Number of Consecutive Terms | The average of the two middle terms. | \( \frac{\text{First Term} + \text{Last Term}}{2} \) |
| Adding Consecutive Terms | Adding \(k\) consecutive terms to \(N\) consecutive terms forms a new longer consecutive sequence. | Average increases by \(k/2\) if added consecutively at the end. |
Additional Information: Arithmetic Progressions
A sequence of consecutive natural numbers is a specific type of arithmetic progression (AP). An arithmetic progression is a sequence of numbers such that the difference between consecutive terms is constant. In the case of consecutive natural numbers, this constant difference is 1.
- The average of any arithmetic progression is the mean of its first and last term.
- If you add terms to an AP such that the resulting sequence is still an AP, the average change can be calculated based on the new first and last terms, or using properties related to adding terms at the end.
- Adding \(k\) terms to the end of an arithmetic progression with common difference \(d\) will shift the average. For consecutive natural numbers, \(d=1\). Adding \(k\) terms consecutively at the end shifts the middle point of the series by \(k/2\), hence increasing the average by \(k/2 \times d = k/2 \times 1 = k/2\).
Understanding arithmetic progressions helps generalize this concept beyond just consecutive natural numbers.
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