Product of a G.P whose `1^(st)`and last term are a and b is P. If two terms equisdistant from both sides are removed the product of remaining terms of series is P', then `P/(P')` is

Options

A. a
B. b
C. `a/b`
D. `a.b`

Answer

To solve the problem, we need to analyze the given information about the geometric progression (G.P.) and the products of its terms.

### Step-by-Step Solution:

1. **Understanding the G.P.**:
Let the first term of the G.P. be \( a \) and the last term be \( b \). The number of terms in the G.P. is \( n \) and the common ratio is \( r \).

2. **Product of the G.P.**:
The product \( P \) of the terms in the G.P. can be expressed as:
\[ P = a^{n/2} \cdot b^{n/2} = (ab)^{n/2} \]
This is because the product of the terms in a G.P. can be represented as the first term raised to the number of terms divided by 2, multiplied by the last term raised to the same power.

3. **Removing Terms**:
When two terms equidistant from both ends are removed, we are left with \( n - 4 \) terms. The remaining terms will still form a G.P. with the first term \( a \) and the last term \( b \).

4. **Product of Remaining Terms**:
The product \( P' \) of the remaining terms can be expressed as:
\[ P' = a^{(n-4)/2} \cdot b^{(n-4)/2} = (ab)^{(n-4)/2} \]

5. **Finding the Ratio \( \frac{P}{P'} \)**:
Now we need to find the ratio \( \frac{P}{P'} \):
\[ \frac{P}{P'} = \frac{(ab)^{n/2}}{(ab)^{(n-4)/2}} \]
This simplifies to:
\[ \frac{P}{P'} = (ab)^{n/2 - (n-4)/2} = (ab)^{(n/2) - (n/2 - 2)} = (ab)^{2} \]

6. **Final Result**:
Therefore, the final result is:
\[ \frac{P}{P'} = (ab)^{2} \]