Enter the sum of squares and the number of consecutive terms to quickly find the matching sequence (consecutive integers, odd numbers, or even numbers).
Finding a sequence of consecutive numbers based on their sum of squares involves some derivation and trial and error. Typically, we use the average method combined with the square root method to estimate the starting number, and then verify through trial. If no solution is found, it can be concluded that no such sequence exists.
The core steps to find the starting number \( a \) are:
Given the sum of squares \( S_{\text{square}} \) and the number of consecutive terms \( n \), we start by assuming that all the consecutive numbers are equal (although this is not the case in reality). Using the square root method, we find an approximate value for the average number. This approximation is generally smaller than the actual starting number.
The average \( \bar{x} \) of these consecutive numbers, whose sum of squares equals the known \( S_{\text{square}} \), can be calculated using the formula: \( \bar{x}^2 \times n = S_{\text{square}} \) Solving for \( \bar{x} \): \( \bar{x} = \sqrt{\frac{S_{\text{square}}}{n}} \) This \( \bar{x} \) is an approximate "average number," typically smaller than the actual starting number \( a \).
Since the consecutive numbers begin at \( a \) and increase from there, the starting number \( a \) can be approximated by the formula: \( a \approx \bar{x} - \frac{n-1}{2} \) This formula gives us a quick way to estimate the starting number.
Once the approximate \( a \) is found, we verify by substituting this value into the sequence formula and calculating the sum of squares. If the sum of squares equals the given \( S_{\text{square}} \), the starting number is correct. If not, we adjust \( a \) by either increasing or decreasing it and try again.
If trial adjustments lead to a sum of squares smaller or larger than the target, it's time to stop testing. Further trials would only increase the error, and we can conclude that no solution exists.
Solution:
1. Calculate the average:
\( \bar{x} = \sqrt{\frac{2135}{7}} = \sqrt{305} \approx 17.46 \)
The average is approximately 17.46.
2. Find the starting odd number:
\( a \approx 17.46 - \frac{7-1}{2} = 17.46 - 3 = 14.46 \)
The approximate starting number is 14.46. Rounding down, we try \( a \approx 15 \).
3. Verification:
\( 15^2 + 17^2 + 19^2 + 21^2 + 23^2 + 25^2 + 27^2 = 225 + 289 + 361 + 441 + 529 + 625 + 729 = 3200 \)
This is larger than 2135, so we reduce \( a \) and try \( a = 13 \).
4. Re-verification:
\( 13^2 + 15^2 + 17^2 + 19^2 + 21^2 + 23^2 + 25^2 = 169 + 225 + 289 + 361 + 441 + 529 + 625 = 2639 \)
Still larger than 2135, so we try \( a = 11 \).
5. Final Verification:
\( 11^2 + 13^2 + 15^2 + 17^2 + 19^2 + 21^2 + 23^2 = 121 + 169 + 225 + 289 + 361 + 441 + 529 = 2135 \)
This matches exactly. Therefore, the 7 consecutive odd numbers are: 11, 13, 15, 17, 19, 21, 23.
Solution:
1. Calculate the average:
\( \bar{x} = \sqrt{\frac{3820}{6}} = \sqrt{636.67} \approx 25.22 \)
The average is approximately 25.22.
2. Find the starting even number:
\( a \approx 25.22 - \frac{6-1}{2} = 25.22 - 2.5 = 22.72 \)
The approximate starting number is 22.72. Rounding down, we try \( a = 22 \).
3. Verification:
\( 22^2 + 24^2 + 26^2 + 28^2 + 30^2 + 32^2 = 484 + 576 + 676 + 784 + 900 + 1024 = 4444 \)
This is larger than 3820, so we reduce \( a \) and try \( a = 20 \).
4. Re-verification:
\( 20^2 + 22^2 + 24^2 + 26^2 + 28^2 + 30^2 = 400 + 484 + 576 + 676 + 784 + 900 = 3820 \)
This matches exactly. Therefore, the 6 consecutive even numbers are: 20, 22, 24, 26, 28, 30.
Solution:
1. Calculate the average:
\( \bar{x} = \sqrt{\frac{925}{2}} = \sqrt{462.5} \approx 21.51 \)
The average is 21.51.
2. Find the starting number:
\( a \approx 21.51 - \frac{2-1}{2} = 21.51 - 0.5 = 21.01 \)
The starting number is \( a = 21 \).
3. Verification:
\( 21^2 + 22^2 = 441 + 484 = 925 \)
This matches exactly. Therefore, the consecutive integers are 21 and 22.
Solution:
1. Calculate the average:
\( \bar{x} = \sqrt{\frac{1060}{2}} = \sqrt{530} \approx 23.02 \)
The average is approximately 23.02.
2. Find the starting even number:
\( a \approx 23.02 - \frac{2-1}{2} = 23.02 - 0.5 = 22.52 \)
The starting number is \( a = 22 \).
3. Verification:
\( 22^2 + 24^2 = 484 + 576 = 1060 \)
This matches exactly. Therefore, the consecutive even numbers are 22 and 24.
Solution:
1. Calculate the average:
\( \bar{x} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5 \)
The average is 5.
2. Find the starting even number:
\( a \approx 5 - \frac{2-1}{2} = 5 - 0.5 = 4.5 \)
The starting number is \( a = 4 \).
3. Verification:
\( 4^2 + 6^2 = 16 + 36 = 52 \)
Since 52 is larger than 50, we try decreasing \( a \), for example, \( a = 2 \).
4. Re-verification:
\( 2^2 + 4^2 = 4 + 16 = 20 \)
This is smaller than 50, and further trials confirm no solution exists.