The question asks to find the value of K in the sequence 1, 2, 3, ..., K, given that the standard deviation of this sequence is 2

To find the value of \( K \) in the sequence \( 1, 2, 3, \ldots, K \), given that the standard deviation of this sequence is 2, follow these steps:

1. **Determine the mean of the sequence:**

   The sequence \( 1, 2, 3, \ldots, K \) is an arithmetic series. The mean \( \mu \) of the first \( K \) positive integers is:
   \[
   \mu = \frac{1 + 2 + 3 + \cdots + K}{K}
   \]
   The sum of the first \( K \) positive integers is given by:
   \[
   \sum_{i=1}^K i = \frac{K(K + 1)}{2}
   \]
   Therefore, the mean \( \mu \) is:
   \[
   \mu = \frac{\frac{K(K + 1)}{2}}{K} = \frac{K + 1}{2}
   \]

2. **Calculate the variance of the sequence:**

   The variance \( \sigma^2 \) of a sequence is the average of the squared differences from the mean. For the sequence \( 1, 2, 3, \ldots, K \), the variance is:
   \[
   \sigma^2 = \frac{1}{K} \sum_{i=1}^K (i - \mu)^2
   \]
   Using the mean \( \mu = \frac{K + 1}{2} \), the squared differences are:
   \[
   \sum_{i=1}^K (i - \mu)^2 = \sum_{i=1}^K \left(i - \frac{K + 1}{2}\right)^2
   \]
   This can be simplified using the formula for the sum of squares of the first \( K \) natural numbers:
   \[
   \sum_{i=1}^K i^2 = \frac{K(K + 1)(2K + 1)}{6}
   \]

3. **Plug in the values and simplify:**

   The variance \( \sigma^2 \) in terms of \( K \) is:
   \[
   \sigma^2 = \frac{1}{K} \sum_{i=1}^K \left(i - \frac{K + 1}{2}\right)^2 = \frac{1}{K} \left(\sum_{i=1}^K i^2 - 2\mu \sum_{i=1}^K i + K\mu^2\right)
   \]
   Substituting \( \sum_{i=1}^K i = \frac{K(K + 1)}{2} \) and \( \sum_{i=1}^K i^2 = \frac{K(K + 1)(2K + 1)}{6} \):
   \[
   \sigma^2 = \frac{1}{K} \left(\frac{K(K + 1)(2K + 1)}{6} - 2\left(\frac{K + 1}{2}\right) \frac{K(K + 1)}{2} + K\left(\frac{K + 1}{2}\right)^2\right)
   \]
   Simplifying the terms inside the parentheses:
   \[
   \sigma^2 = \frac{1}{K} \left(\frac{K(K + 1)(2K + 1)}{6} - \frac{K(K + 1)^2}{2} + \frac{K(K + 1)^2}{4}\right)
   \]
   Combining like terms and simplifying further:
   \[
   \sigma^2 = \frac{1}{K} \left(\frac{K(K + 1)}{12} (2K + 1 - 6 + 3)\right)
   \]
   \[
   \sigma^2 = \frac{1}{K} \left(\frac{K(K + 1)}{12} (2K - 2)\right) = \frac{1}{K} \left(\frac{K(K + 1)(K - 1)}{6}\right)
   \]
   \[
   \sigma^2 = \frac{(K + 1)(K - 1)}{6} = \frac{K^2 - 1}{6}
   \]

4. **Set the variance equal to 4 (since the standard deviation is 2) and solve for \( K \):**

   \[
   \frac{K^2 - 1}{6} = 4
   \]
   \[
   K^2 - 1 = 24
   \]
   \[
   K^2 = 25
   \]
   \[
   K = 5
   \]

Thus, the value of \( K \) is \( 5 \).

Would you like more details on any part of this solution, or do you have any questions?

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Here are some related questions to deepen your understanding:

1. What is the formula for the sum of the first \( n \) natural numbers?
2. How do you derive the formula for the sum of the squares of the first \( n \) natural numbers?
3. What is the difference between variance and standard deviation?
4. How does the variance of a sequence change if you add a constant to each term in the sequence?
5. How would you find the standard deviation of a sequence if you were given the variance?
6. Can the standard deviation of a sequence ever be negative? Why or why not?
7. How does the mean of a sequence of consecutive integers relate to the range of the sequence?
8. If the sequence were \( 2, 3, 4, \ldots, K+1 \), how would the mean and standard deviation change?

**Tip:** Always double-check your calculations, especially when working with sums and variances, as small errors can lead to incorrect results.

Discover how to find the value of K in the sequence 1 to K, where the standard deviation of the sequence is 2. Learn about arithmetic series, variance, and the steps to calculate K using mathematical formulas. Suitable for high school students and beyond.

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