STATEMENT-1 : The sum of 100 arithmetic means between two given number 1000 and 3016 is 200800 and
STATEMENT-2 : The sum of n arithmetic means between two given numbers is ` n^(th)` power of their single A.M.

Options

A. Statemant-1 is True , Statement-2 is True, Statement -2 is a correct explanation for Statement-1
B. Statemant-1 is True , Statement-2 is True, Statement -2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Stetement-2 is False.
D. Statement-1 is False, Statement-2 is True

Answer

To solve the given problem, we need to analyze both statements provided and determine their validity.

### Step-by-Step Solution:

**Step 1: Analyze Statement 1**

The statement claims that the sum of 100 arithmetic means between the numbers 1000 and 3016 is 200800.

To find the sum of the arithmetic means, we first need to understand how many terms we are dealing with. We have:
- First term (a) = 1000
- Last term (b) = 3016
- Number of arithmetic means (n) = 100

The total number of terms (including the two given numbers) will be:
\[ \text{Total terms} = n + 2 = 100 + 2 = 102 \]

The formula for the sum of an arithmetic series is:
\[ S = \frac{n}{2} \times (\text{first term} + \text{last term}) \]

In our case, the sum of all 102 terms will be:
\[ S = \frac{102}{2} \times (1000 + 3016) \]
\[ S = 51 \times 4016 \]
\[ S = 204816 \]

However, we are interested in the sum of the 100 arithmetic means only. Thus, we need to subtract the first and last terms from the total sum:
\[ \text{Sum of 100 arithmetic means} = S - (1000 + 3016) \]
\[ = 204816 - 4016 \]
\[ = 200800 \]

So, Statement 1 is **True**.

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**Step 2: Analyze Statement 2**

The statement claims that the sum of n arithmetic means between two given numbers is the nth power of their single arithmetic mean.

Let’s denote the two numbers as \( a \) and \( b \). The arithmetic mean (A.M.) of these two numbers is given by:
\[ A.M. = \frac{a + b}{2} \]

The sum of n arithmetic means inserted between \( a \) and \( b \) can be calculated as follows:
- The total number of terms is \( n + 2 \) (including \( a \) and \( b \)).
- The sum of these \( n + 2 \) terms is:
\[ S = \frac{n + 2}{2} \times (a + b) \]

To find the sum of just the n arithmetic means, we need to subtract \( a \) and \( b \):
\[ \text{Sum of n arithmetic means} = S - (a + b) \]
\[ = \frac{n + 2}{2} \times (a + b) - (a + b) \]
\[ = (a + b) \left( \frac{n + 2}{2} - 1 \right) \]
\[ = (a + b) \left( \frac{n}{2} \right) \]

This shows that the sum of n arithmetic means is proportional to \( a + b \) and not the nth power of the arithmetic mean.

Thus, Statement 2 is **False**.

### Conclusion:

- **Statement 1** is True.
- **Statement 2** is False.