If `a_(1),a_(2),a_(3)"....."` are in GP with first term a and common ratio r, then `(a_(1)a_(2))/(a_(1)^(2)-a_(2)^(2))+(a_(2)a_(3))/(a_(2)^(2)-a_(3)^(2))+(a_(3)a_(4))/(a_(3)^(2)-a_(4)^(2))+"....."+(a_(n-1)a_(n))/(a_(n-1)^(2)-a_(n)^(2))` is equal to

Options

A. `(nr)/(1-r^(2))`
B. `((n-1)r)/(1-r^(2))`
C. `(nr)/(1-r)`
D. `((n-1)r)/(1-r)`

Explanation

To solve the problem, we start with the given terms in a geometric progression (GP). The first term is \( a \) and the common ratio is \( r \). Thus, we can express the terms as follows:

- \( a_1 = a \)
- \( a_2 = ar \)
- \( a_3 = ar^2 \)
- \( a_4 = ar^3 \)
- ...
- \( a_n = ar^{n-1} \)

Now, we need to evaluate the expression:

\[ \frac{a_1 a_2}{a_1^2 - a_2^2} + \frac{a_2 a_3}{a_2^2 - a_3^2} + \frac{a_3 a_4}{a_3^2 - a_4^2} + \ldots + \frac{a_{n-1} a_n}{a_{n-1}^2 - a_n^2} \]

### Step 1: Substitute the terms in the expression

Substituting the terms into the expression, we have:

\[ \frac{a \cdot ar}{a^2 - (ar)^2} + \frac{(ar) \cdot (ar^2)}{(ar)^2 - (ar^2)^2} + \frac{(ar^2) \cdot (ar^3)}{(ar^2)^2 - (ar^3)^2} + \ldots + \frac{(ar^{n-2}) \cdot (ar^{n-1})}{(ar^{n-2})^2 - (ar^{n-1})^2} \]

### Step 2: Simplify each term

For the first term:

\[ \frac{a^2 r}{a^2 - a^2 r^2} = \frac{a^2 r}{a^2(1 - r^2)} = \frac{r}{1 - r^2} \]

For the second term:

\[ \frac{(ar)(ar^2)}{(ar)^2 - (ar^2)^2} = \frac{a^2 r^3}{a^2 r^2(1 - r^2)} = \frac{r^3}{r^2(1 - r^2)} = \frac{r}{1 - r^2} \]

Continuing this process, we find that each term simplifies to:

\[ \frac{r}{1 - r^2} \]

### Step 3: Count the number of terms

The number of terms in the sum is \( n - 1 \) since we are summing from \( a_1 \) to \( a_{n-1} \).

### Step 4: Combine the terms

Thus, the entire expression simplifies to:

\[ (n - 1) \cdot \frac{r}{1 - r^2} \]

### Final Answer

The final result is:

\[ \frac{(n - 1)r}{1 - r^2} \]

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