If the `(p+q)^(th)` term of a G.P. is `a` and `(p-q)^(th)` term is `b`, determine its `p^(th)` term.

Options

A. mm
B. `(m)/(n)`
C. `sqrt(mn)`
D. `sqrt((m)/(n))`

Answer

To find the \( p^{th} \) term of a geometric progression (G.P.) given that the \( (p+q)^{th} \) term is \( a \) and the \( (p-q)^{th} \) term is \( b \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the General Term of G.P.**:
The \( n^{th} \) term of a G.P. can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

2. **Expressing the Given Terms**:
For the \( (p+q)^{th} \) term:
\[ T_{p+q} = a \cdot r^{(p+q)-1} = a \]
For the \( (p-q)^{th} \) term:
\[ T_{p-q} = a \cdot r^{(p-q)-1} = b \]

3. **Setting Up the Equations**:
From the above expressions, we can write two equations:
\[ a \cdot r^{p+q-1} = a \quad \text{(1)} \]
\[ a \cdot r^{p-q-1} = b \quad \text{(2)} \]

4. **Dividing the Equations**:
We can divide equation (1) by equation (2):
\[ \frac{a \cdot r^{p+q-1}}{a \cdot r^{p-q-1}} = \frac{a}{b} \]
This simplifies to:
\[ r^{(p+q-1) - (p-q-1)} = \frac{a}{b} \]
Simplifying the exponent:
\[ r^{(p+q-1) - (p-q-1)} = r^{2q} = \frac{a}{b} \]

5. **Finding the Common Ratio**:
Taking the square root of both sides:
\[ r^q = \sqrt{\frac{a}{b}} \]

6. **Finding the \( p^{th} \) Term**:
Now, we can find the \( p^{th} \) term:
\[ T_p = a \cdot r^{p-1} \]
Substituting \( r^{q} \):
\[ r^{p-1} = r^{q} \cdot r^{(p-1)-q} = \sqrt{\frac{a}{b}} \cdot r^{p-q-1} \]
We need to express \( r^{p-q-1} \) in terms of \( a \) and \( b \). From equation (2):
\[ r^{p-q-1} = \frac{b}{a} \]
Thus:
\[ T_p = a \cdot \left(\sqrt{\frac{a}{b}} \cdot \frac{b}{a}\right) = a \cdot \sqrt{\frac{a}{b}} \cdot \frac{b}{a} = \sqrt{ab} \]

### Final Result:
The \( p^{th} \) term of the G.P. is:
\[ \boxed{\sqrt{ab}} \]