If `barx`is at mean of n values of x, then `Sigma_(i=1)^(n) (x_(i)- barx)=0` and if a has value other than `barx " then " Sigma_(i=1)^(n) (x_(i)- barx)^(2) " is less than " Sigma(x_(i)-a)^(2)`

Explanation

To solve the given question step by step, let's break it down into two parts as stated in the question.

### Part 1: Proving that \( \Sigma_{i=1}^{n} (x_i - \bar{x}) = 0 \)

1. **Definition of Mean**: The mean \( \bar{x} \) of \( n \) values \( x_1, x_2, \ldots, x_n \) is defined as:
\[ \bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} \]

2. **Rearranging the Mean**: From the definition, we can express the sum of the values as:
\[ x_1 + x_2 + \ldots + x_n = n \bar{x} \]

3. **Summation Expression**: Now, we need to evaluate the expression \( \Sigma_{i=1}^{n} (x_i - \bar{x}) \):
\[ \Sigma_{i=1}^{n} (x_i - \bar{x}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + \ldots + (x_n - \bar{x}) \]

4. **Factoring Out the Mean**: This can be rewritten as:
\[ = (x_1 + x_2 + \ldots + x_n) - n\bar{x} \]

5. **Substituting the Sum**: Substitute \( x_1 + x_2 + \ldots + x_n \) with \( n\bar{x} \):
\[ = n\bar{x} - n\bar{x} = 0 \]

Thus, we have shown that:
\[ \Sigma_{i=1}^{n} (x_i - \bar{x}) = 0 \]

### Part 2: Proving that \( \Sigma_{i=1}^{n} (x_i - \bar{x})^2 < \Sigma_{i=1}^{n} (x_i - a)^2 \) for \( a \neq \bar{x} \)

1. **Expanding the Squares**: We need to expand both sides of the inequality:
- Left-hand side:
\[ \Sigma_{i=1}^{n} (x_i - \bar{x})^2 \]
- Right-hand side:
\[ \Sigma_{i=1}^{n} (x_i - a)^2 \]

2. **Using the Expansion Formula**: The expansion of \( (x_i - a)^2 \) gives:
\[ (x_i - a)^2 = (x_i - \bar{x} + \bar{x} - a)^2 \]
Expanding this, we get:
\[ = (x_i - \bar{x})^2 + 2(x_i - \bar{x})(\bar{x} - a) + (\bar{x} - a)^2 \]

3. **Summing Over All i**: Now, summing this over \( i \):
\[ \Sigma_{i=1}^{n} (x_i - a)^2 = \Sigma_{i=1}^{n} (x_i - \bar{x})^2 + 2(\bar{x} - a) \Sigma_{i=1}^{n} (x_i - \bar{x}) + n(\bar{x} - a)^2 \]

4. **Using the Result from Part 1**: From Part 1, we know that \( \Sigma_{i=1}^{n} (x_i - \bar{x}) = 0 \). Therefore, the second term vanishes:
\[ \Sigma_{i=1}^{n} (x_i - a)^2 = \Sigma_{i=1}^{n} (x_i - \bar{x})^2 + n(\bar{x} - a)^2 \]

5. **Final Inequality**: Since \( n(\bar{x} - a)^2 > 0 \) when \( a \neq \bar{x} \), we conclude:
\[ \Sigma_{i=1}^{n} (x_i - \bar{x})^2 < \Sigma_{i=1}^{n} (x_i - a)^2 \]

Thus, we have shown that:
\[ \Sigma_{i=1}^{n} (x_i - \bar{x})^2 < \Sigma_{i=1}^{n} (x_i - a)^2 \text{ for } a \neq \bar{x} \]

### Summary of Results
1. \( \Sigma_{i=1}^{n} (x_i - \bar{x}) = 0 \)
2. \( \Sigma_{i=1}^{n} (x_i - \bar{x})^2 < \Sigma_{i=1}^{n} (x_i - a)^2 \) for \( a \neq \bar{x} \)