2, p, q are in G.P. (where p ne q) and in A.P., 2 is third term, p is 7th term and q is 8th term, the `64(p^2 + q^2)`

Options

A. 17
B. 18
C. 19
D. 20

Answer

To solve the problem, we need to find the value of \( 64(p^2 + q^2) \) given that \( 2, p, q \) are in both geometric progression (G.P.) and arithmetic progression (A.P.), with specific terms defined.

### Step-by-Step Solution:

1. **Understanding the Progressions**:
- Since \( 2, p, q \) are in G.P., we can express \( p \) and \( q \) in terms of a common ratio \( r \):
\[ p = 2r \quad \text{and} \quad q = 2r^2 \]

2. **Setting Up the A.P.**:
- In an A.P., the terms can be expressed as:
- 3rd term: \( a + 2d = 2 \)
- 7th term: \( a + 6d = p \)
- 8th term: \( a + 7d = q \)

3. **Expressing Terms**:
- From the A.P. definitions, we have:
\[ a + 2d = 2 \quad \text{(1)} \]
\[ a + 6d = p \quad \text{(2)} \]
\[ a + 7d = q \quad \text{(3)} \]

4. **Subtracting Equations**:
- Subtract equation (1) from (2):
\[ (a + 6d) - (a + 2d) = p - 2 \implies 4d = p - 2 \quad \text{(4)} \]
- Subtract equation (2) from (3):
\[ (a + 7d) - (a + 6d) = q - p \implies d = q - p \quad \text{(5)} \]

5. **Substituting for \( d \)**:
- From (4), we can express \( d \) in terms of \( p \):
\[ d = \frac{p - 2}{4} \quad \text{(6)} \]
- From (5), we can express \( d \) in terms of \( q \) and \( p \):
\[ d = q - p \quad \text{(7)} \]

6. **Equating the Two Expressions for \( d \)**:
- Set equations (6) and (7) equal:
\[ \frac{p - 2}{4} = q - p \]
- Multiply through by 4:
\[ p - 2 = 4(q - p) \implies p - 2 = 4q - 4p \]
- Rearranging gives:
\[ 5p - 4q = 2 \quad \text{(8)} \]

7. **Substituting for \( p \) and \( q \)**:
- Substitute \( p = 2r \) and \( q = 2r^2 \) into equation (8):
\[ 5(2r) - 4(2r^2) = 2 \implies 10r - 8r^2 = 2 \]
- Rearranging gives:
\[ 8r^2 - 10r + 2 = 0 \]

8. **Solving the Quadratic Equation**:
- Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 8 \cdot 2}}{2 \cdot 8} = \frac{10 \pm \sqrt{100 - 64}}{16} = \frac{10 \pm \sqrt{36}}{16} \]
\[ r = \frac{10 \pm 6}{16} \]
- This gives us:
\[ r = 1 \quad \text{or} \quad r = \frac{1}{4} \]

9. **Finding \( p \) and \( q \)**:
- If \( r = 1 \), then \( p = 2 \) and \( q = 2 \) which contradicts \( p \neq q \).
- Therefore, \( r = \frac{1}{4} \):
\[ p = 2 \cdot \frac{1}{4} = \frac{1}{2}, \quad q = 2 \cdot \left(\frac{1}{4}\right)^2 = \frac{1}{8} \]

10. **Calculating \( 64(p^2 + q^2) \)**:
- Calculate \( p^2 + q^2 \):
\[ p^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}, \quad q^2 = \left(\frac{1}{8}\right)^2 = \frac{1}{64} \]
\[ p^2 + q^2 = \frac{1}{4} + \frac{1}{64} = \frac{16}{64} + \frac{1}{64} = \frac{17}{64} \]
- Now multiply by 64:
\[ 64(p^2 + q^2) = 64 \cdot \frac{17}{64} = 17 \]

### Final Answer:
Thus, the value of \( 64(p^2 + q^2) \) is \( \boxed{17} \).