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, let's break it down step by step.

### Step 1: Understand the GP
We are given a geometric progression (G.P.) with the first term \( a \) and the last term \( b \). The terms of the G.P. can be expressed as:
\[ a, ar, ar^2, ar^3, \ldots, ar^{n-1} \]
where \( r \) is the common ratio and \( n \) is the total number of terms.

### Step 2: Calculate the Product of the G.P.
The product \( P \) of all terms in the G.P. can be calculated as follows:
\[ P = a \cdot ar \cdot ar^2 \cdots ar^{n-1} = a^n \cdot r^{\frac{(n-1)n}{2}} \]
However, since we know that the first term is \( a \) and the last term is \( b \), we can express \( b \) as:
\[ b = ar^{n-1} \]
Thus, the product \( P \) can also be expressed as:
\[ P = (ab)^{\frac{n}{2}} \]

### Step 3: Removing Terms from the G.P.
Next, we need to consider what happens when we remove two terms that are equidistant from both ends of the G.P. This means we remove the first term \( a \) and the last term \( b \).

### Step 4: Calculate the Product of Remaining Terms
After removing \( a \) and \( b \), the remaining terms of the G.P. would be:
\[ ar, ar^2, ar^3, \ldots, ar^{n-2} \]
The number of remaining terms is \( n - 2 \). The product \( P' \) of the remaining terms can be calculated as:
\[ P' = ar \cdot ar^2 \cdots ar^{n-2} = a^{n-2} \cdot r^{\frac{(n-2)(n-1)}{2}} \]
Using the relationship between \( a \), \( b \), and \( r \), we can express \( P' \) as:
\[ P' = (ab)^{\frac{n-2}{2}} \]

### Step 5: Calculate \( \frac{P}{P'} \)
Now, we can find the ratio \( \frac{P}{P'} \):
\[ \frac{P}{P'} = \frac{(ab)^{\frac{n}{2}}}{(ab)^{\frac{n-2}{2}}} \]
This simplifies to:
\[ \frac{P}{P'} = (ab)^{\frac{n}{2} - \frac{n-2}{2}} = (ab)^{1} = ab \]

### Final Answer
Thus, the value of \( \frac{P}{P'} \) is:
\[ \frac{P}{P'} = ab \]