Consider the following statements:
1. The algebraic sum of deviations of a set of values from their arithmetic mean is always zero.
2. Arithmetic mean gt Median gt Mode for a symmetric distribution.
Which of the above statements is/are correct?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To solve the question, we need to evaluate the validity of the two given statements regarding statistics.

### Step 1: Analyze the First Statement
The first statement is: "The algebraic sum of deviations of a set of values from their arithmetic mean is always zero."

**Solution:**
1. Let the set of values be \( a_1, a_2, a_3, \ldots, a_n \).
2. The arithmetic mean (average) of these values is calculated as:
\[ \text{Mean} = \frac{a_1 + a_2 + a_3 + \ldots + a_n}{n} \]
3. The deviation of each value from the mean is given by:
\[ d_i = a_i - \text{Mean} \]
4. Therefore, the algebraic sum of deviations is:
\[ \sum_{i=1}^{n} d_i = \sum_{i=1}^{n} (a_i - \text{Mean}) = \sum_{i=1}^{n} a_i - n \cdot \text{Mean} \]
5. Substituting the expression for the mean, we have:
\[ \sum_{i=1}^{n} a_i - n \cdot \frac{a_1 + a_2 + \ldots + a_n}{n} = \sum_{i=1}^{n} a_i - \sum_{i=1}^{n} a_i = 0 \]
6. Thus, the algebraic sum of deviations from the mean is indeed zero.

**Conclusion:** The first statement is **true**.

### Step 2: Analyze the Second Statement
The second statement is: "Arithmetic mean > Median > Mode for a symmetric distribution."

**Solution:**
1. In a symmetric distribution, the mean, median, and mode are all equal.
2. Therefore, for a symmetric distribution:
\[ \text{Mean} = \text{Median} = \text{Mode} \]
3. This contradicts the statement that states the mean is greater than the median, which is also greater than the mode.

**Conclusion:** The second statement is **false**.

### Final Conclusion
- The first statement is true.
- The second statement is false.

Thus, the correct answer is that only the first statement is correct.