Statement -1: There exists no A.P. whose three terms are `sqrt(3),sqrt(5)andsqrt(7)`.
Statement-2: If `a_(p),a_(q)anda_(r)` are three distinct terms of an A.P., then `(a_(p)-a_(q))/(a_(p)-q_(r))` is a rational number.

Options

A. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.
B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.
C. Statement -1 is true, Statement -2 is False.
D. Statement -1 is False, Statement -2 is True.

Explanation

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

### Step 1: Analyze Statement 1
**Statement 1:** There exists no A.P. whose three terms are \( \sqrt{3}, \sqrt{5}, \text{ and } \sqrt{7} \).

To prove this statement, we will assume that there exists an arithmetic progression (A.P.) with these three terms.

Let the three terms be \( a, b, c \) where:
- \( a = \sqrt{3} \)
- \( b = \sqrt{5} \)
- \( c = \sqrt{7} \)

In an A.P., the middle term \( b \) must be the average of the other two terms:
\[ b = \frac{a + c}{2} \]

Substituting the values:
\[ \sqrt{5} = \frac{\sqrt{3} + \sqrt{7}}{2} \]

Multiplying both sides by 2 gives:
\[ 2\sqrt{5} = \sqrt{3} + \sqrt{7} \]

### Step 2: Square Both Sides
Now, we will square both sides to eliminate the square roots:
\[ (2\sqrt{5})^2 = (\sqrt{3} + \sqrt{7})^2 \]
\[ 4 \cdot 5 = 3 + 7 + 2\sqrt{21} \]
\[ 20 = 10 + 2\sqrt{21} \]

### Step 3: Isolate the Square Root
Now, isolate the square root term:
\[ 20 - 10 = 2\sqrt{21} \]
\[ 10 = 2\sqrt{21} \]
\[ 5 = \sqrt{21} \]

### Step 4: Square Again
Square both sides again:
\[ 5^2 = 21 \]
\[ 25 = 21 \]

This is a contradiction, which means our assumption that \( \sqrt{3}, \sqrt{5}, \text{ and } \sqrt{7} \) can form an A.P. is false.

**Conclusion for Statement 1:** Therefore, Statement 1 is true.

### Step 5: Analyze Statement 2
**Statement 2:** If \( a_p, a_q, a_r \) are three distinct terms of an A.P., then \( \frac{a_p - a_q}{a_p - a_r} \) is a rational number.

Let’s denote the terms of the A.P. as:
\[ a_p = a + (p-1)d, \quad a_q = a + (q-1)d, \quad a_r = a + (r-1)d \]

Now, compute \( a_p - a_q \) and \( a_p - a_r \):
\[ a_p - a_q = (a + (p-1)d) - (a + (q-1)d) = (p-q)d \]
\[ a_p - a_r = (a + (p-1)d) - (a + (r-1)d) = (p-r)d \]

Now, substitute these into the expression:
\[ \frac{a_p - a_q}{a_p - a_r} = \frac{(p-q)d}{(p-r)d} = \frac{p-q}{p-r} \]

Since \( p, q, r \) are distinct integers, \( p-q \) and \( p-r \) are both integers, making their ratio a rational number.

**Conclusion for Statement 2:** Therefore, Statement 2 is also true.

### Final Conclusion
Both statements are true:
- Statement 1 is true: There exists no A.P. whose three terms are \( \sqrt{3}, \sqrt{5}, \text{ and } \sqrt{7} \).
- Statement 2 is true: If \( a_p, a_q, a_r \) are three distinct terms of an A.P., then \( \frac{a_p - a_q}{a_p - a_r} \) is a rational number.