Enter the sum and count to quickly find the corresponding consecutive numbers (integers, odd numbers, or even numbers).
When you know the sum and count of a series of consecutive numbers, you can actually use mathematical formulas to derive these numbers. Different types of consecutive numbers (integers, odd numbers, or even numbers) have their own rules and formulas. Below are the detailed steps for manually deriving each case.
For a set of consecutive integers (an arithmetic sequence with a common difference of 1), let the sum be \( S \) and the count be \( n \). You can find the starting number \( a \) and the ending number \( b \) using the following steps.
\( S = \frac{n \times (a + b)}{2} \) Since \( b = a + n - 1 \), substitute into the formula: \( S = \frac{n \times (a + a + n - 1)}{2} = \frac{n \times (2a + n - 1)}{2} \) Simplifying gives: \( S = \frac{n \times (2a + n - 1)}{2} \) Solve for the starting number \( a \): \( a = \frac{2S - n(n - 1)}{2n} \) Once you have \( a \), the ending number \( b \) is \( b = a + n - 1 \).
Solution:
Substitute the sum and count into the formula:
\( a = \frac{2 \times 55 - 5 \times (5 - 1)}{2 \times 5} = \frac{110 - 20}{10} = 9 \)
The starting number is \( a = 9 \), and the ending number is \( b = 9 + 5 - 1 = 14 \).
Result: The consecutive integers are 9, 10, 11, 12, and 13.
For a set of consecutive odd numbers (also an arithmetic sequence with a common difference of 2), let the sum be \( S \) and the count be \( n \). The starting odd number \( a \) and ending odd number \( b \) can be derived similarly.
\( S = \frac{n \times (a + b)}{2} \) where \( b = a + 2(n - 1) \), substitute into the formula: \( S = \frac{n \times (a + a + 2(n - 1))}{2} = \frac{n \times (2a + 2(n - 1))}{2} \) Simplifying: \( S = n \times (a + n - 1) \) Solve for the starting odd number \( a \): \( a = \frac{S}{n} - (n - 1) \) Then the ending odd number \( b \) is \( b = a + 2(n - 1) \).
Solution:
Substitute the data into the formula:
\( a = \frac{25}{5} - (5 - 1) = 5 - 4 = 1\)
The starting odd number is \( a = 1 \), and the ending odd number is \( b = 1 + 2(5 - 1) = 9 \).
Result: The consecutive odd numbers are 1, 3, 5, 7, and 9.
For a set of consecutive even numbers, the derivation process is similar to that of consecutive odd numbers.
\( S = \frac{n \times (a + b)}{2} \) where \( b = a + 2(n - 1) \), substitute into the formula: \( S = n \times (a + n - 1) \) Solve for the starting even number \( a \): \( a = \frac{S}{n} - (n - 1) \) The ending even number \( b \) is \( b = a + 2(n - 1) \). This formula is identical to that for consecutive odd numbers.
Solution:
Substitute the data into the formula:
\( a = \frac{30}{5} - (5 - 1) = 6 - 4 = 2\)
The starting even number is \( a = 2 \), and the ending even number is \( b = 2 + 2(5 - 1) = 10 \).
The starting even number is \( a = 2 \), and the ending even number is \( b = 2 + 2(5 - 1) = 10 \).