If there be n quantities in G.P., whose common ratio is r and `S_(m)` denotes the sum of the first m terms, then the sum of their products, taken two by two, is

Options

A. `S_(m)S_(m-1)`
B. `(r)/(r+1)S_(m)S_(m-1)`
C. `(r)/(r-1)S_(m)S_(m-1)`
D. `(r+1)/(r)S_(m)S_(m-1)`

Answer

To solve the problem of finding the sum of the products of n quantities in a geometric progression (G.P.) taken two at a time, we can follow these steps:

### Step 1: Define the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The terms of the G.P. can be expressed as:
\[ a, ar, ar^2, ar^3, \ldots, ar^{n-1} \]

### Step 2: Sum of the first m terms
The sum of the first \( m \) terms of a G.P. is given by the formula:
\[ S_m = a \frac{r^m - 1}{r - 1} \]

### Step 3: Find the sum of products taken two at a time
We need to find the sum of the products of the terms taken two at a time, which can be represented as:
\[ \sum_{i=1}^{n} a_i a_j \quad \text{for } i \neq j \]

This can be expressed in terms of the square of the sum of the terms:
\[ \left( \sum_{i=1}^{n} a_i \right)^2 = \sum_{i=1}^{n} a_i^2 + 2 \sum_{i < j} a_i a_j \]
From this, we can isolate the sum of products:
\[ \sum_{i < j} a_i a_j = \frac{1}{2} \left( \left( \sum_{i=1}^{n} a_i \right)^2 - \sum_{i=1}^{n} a_i^2 \right) \]

### Step 4: Calculate \( \sum_{i=1}^{n} a_i \)
Using the formula for the sum of the first \( m \) terms:
\[ \sum_{i=1}^{m} a_i = S_m = a \frac{r^m - 1}{r - 1} \]

### Step 5: Calculate \( \sum_{i=1}^{n} a_i^2 \)
The sum of the squares of the terms in the G.P. is:
\[ \sum_{i=1}^{m} a_i^2 = a^2 + (ar)^2 + (ar^2)^2 + \ldots + (ar^{m-1})^2 = a^2 (1 + r^2 + r^4 + \ldots + r^{2(m-1)}) \]
This is also a G.P. with first term \( a^2 \) and common ratio \( r^2 \):
\[ \sum_{i=1}^{m} a_i^2 = a^2 \frac{r^{2m} - 1}{r^2 - 1} \]

### Step 6: Substitute back into the equation
Now substituting \( S_m \) and \( \sum_{i=1}^{m} a_i^2 \) into the equation for \( \sum_{i < j} a_i a_j \):
\[ \sum_{i < j} a_i a_j = \frac{1}{2} \left( \left( S_m \right)^2 - \sum_{i=1}^{m} a_i^2 \right) \]
\[ = \frac{1}{2} \left( \left( a \frac{r^m - 1}{r - 1} \right)^2 - a^2 \frac{r^{2m} - 1}{r^2 - 1} \right) \]

### Step 7: Final expression
This simplifies to:
\[ = \frac{1}{2} \left( \frac{a^2 (r^m - 1)^2}{(r - 1)^2} - a^2 \frac{r^{2m} - 1}{r^2 - 1} \right) \]

### Summary
The sum of the products of the terms taken two at a time is:
\[ \sum_{i < j} a_i a_j = \frac{1}{2} \left( \frac{a^2 (r^m - 1)^2}{(r - 1)^2} - a^2 \frac{r^{2m} - 1}{r^2 - 1} \right) \]