In a G.P. if the `(m + n)^(th)` term be p and (m - n)th term be q, then its `m^(th)` term is

Options

A. `sqrt (pq)`
B. `sqrt(p//q)`
C. `sqrt(q//p)`
D. `p//q`

Answer

To solve the problem, we need to find the \( m^{th} \) term of a geometric progression (G.P.) given that the \( (m+n)^{th} \) term is \( p \) and the \( (m-n)^{th} \) term is \( q \).

### Step-by-Step Solution:

1. **Define the Terms of the G.P.:**
Let the first term of the G.P. be \( a \) and the common ratio be \( r \).
The \( n^{th} \) term of a G.P. is given by:
\[ T_n = a r^{n-1} \]

2. **Write the Equations for Given Terms:**
The \( (m+n)^{th} \) term is:
\[ T_{m+n} = a r^{(m+n)-1} = p \quad \text{(1)} \]
The \( (m-n)^{th} \) term is:
\[ T_{m-n} = a r^{(m-n)-1} = q \quad \text{(2)} \]

3. **Divide Equation (1) by Equation (2):**
To eliminate \( a \), divide the first equation by the second:
\[ \frac{T_{m+n}}{T_{m-n}} = \frac{p}{q} \]
This gives:
\[ \frac{a r^{m+n-1}}{a r^{m-n-1}} = \frac{p}{q} \]
Simplifying this, we have:
\[ r^{(m+n-1) - (m-n-1)} = \frac{p}{q} \]
This simplifies to:
\[ r^{2n} = \frac{p}{q} \]

4. **Solve for \( r \):**
Taking the square root of both sides:
\[ r^n = \sqrt{\frac{p}{q}} \quad \text{(3)} \]

5. **Find the \( m^{th} \) Term:**
Now, we need to find the \( m^{th} \) term \( T_m \):
\[ T_m = a r^{m-1} \]
We can express \( a \) in terms of \( p \) using equation (1):
\[ a = \frac{p}{r^{m+n-1}} \quad \text{(4)} \]

6. **Substitute \( a \) and \( r \) into \( T_m \):**
Substitute equation (4) into the expression for \( T_m \):
\[ T_m = \frac{p}{r^{m+n-1}} \cdot r^{m-1} \]
This simplifies to:
\[ T_m = p \cdot r^{(m-1) - (m+n-1)} = p \cdot r^{-n} \]
Now substitute \( r^n \) from equation (3):
\[ T_m = p \cdot \frac{1}{\sqrt{\frac{p}{q}}} = p \cdot \sqrt{\frac{q}{p}} = \sqrt{pq} \]

### Final Answer:
Thus, the \( m^{th} \) term of the G.P. is:
\[ \boxed{\sqrt{pq}} \]