If in a g.P. `{ t_(n))` it is given that `t_(p+q)` =a and `t_(p-q)` = b then : `t_(p)`= (A) `(ab)^(1/2)` (B)`(ab)^(1/3)` (C)`(ab)^(1/4)` (D) none of these

Options

A. `(ab)^(1//2)`
B. `(ab)^(1//3)`
C. `(ab)^(1//4)`
D. none of these

Answer

To solve the problem, we need to find the term \( t_p \) in a geometric progression (G.P.) given that \( t_{p+q} = a \) and \( t_{p-q} = b \).

### Step-by-Step Solution:

1. **Understanding the Terms of G.P.**:
In a geometric progression, the \( n \)-th term can be expressed as:
\[ t_n = a_1 \cdot r^{n-1} \]
where \( a_1 \) is the first term and \( r \) is the common ratio.

2. **Expressing Given Terms**:
From the problem, we have:
\[ t_{p+q} = a_1 \cdot r^{(p+q)-1} = a \]
\[ t_{p-q} = a_1 \cdot r^{(p-q)-1} = b \]

3. **Setting Up the Equations**:
We can rewrite the equations as:
\[ a_1 \cdot r^{p+q-1} = a \quad \text{(1)} \]
\[ a_1 \cdot r^{p-q-1} = b \quad \text{(2)} \]

4. **Multiplying the Two Equations**:
Multiply equations (1) and (2):
\[ (a_1 \cdot r^{p+q-1}) \cdot (a_1 \cdot r^{p-q-1}) = a \cdot b \]
This simplifies to:
\[ a_1^2 \cdot r^{(p+q-1) + (p-q-1)} = ab \]
\[ a_1^2 \cdot r^{2p-2} = ab \]

5. **Rearranging the Equation**:
We can rearrange this to find \( a_1 \):
\[ a_1^2 = ab \cdot r^{2 - 2p} \]

6. **Finding \( t_p \)**:
Now, we need to find \( t_p \):
\[ t_p = a_1 \cdot r^{p-1} \]
Substituting \( a_1 \) from the previous step:
\[ t_p = \sqrt{ab \cdot r^{2 - 2p}} \cdot r^{p-1} \]
Simplifying this gives:
\[ t_p = \sqrt{ab} \cdot r^{(2 - 2p + p - 1)} = \sqrt{ab} \cdot r^{1 - p} \]

7. **Finding the Value of \( r \)**:
To express \( t_p \) purely in terms of \( a \) and \( b \), we can assume \( r = 1 \) (which is valid since \( r \) does not affect the ratio):
\[ t_p = \sqrt{ab} \]

8. **Final Answer**:
Thus, we conclude that:
\[ t_p = (ab)^{1/2} \]
Therefore, the correct option is:
\[ \text{(A) } (ab)^{1/2} \]