If `a_1,a_2,a_3, b, ... ,a_n` are an A.P. of non-zero terms, prove that `1/(a_1a_2)+1/(a_2a_3)+ ... +1/(a_(n-1)a_n)=`

Options

A. `(n-1)/(a_1a_n)dot`
B. `(n+1)/(a_1a_n)dot`
C. `(1-n)/(a_1a_n)dot`
D. `(n)/(a_1a_n)dot`

Explanation

To prove the given statement, we need to show that:

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

where \( a_1, a_2, a_3, \ldots, a_n \) are terms of an arithmetic progression (A.P.) with non-zero terms.

### Step-by-Step Solution:

1. **Identify the terms of the A.P.**:
Let the first term be \( a_1 \) and the common difference be \( d \). Then the terms can be expressed as:
\[ a_2 = a_1 + d, \quad a_3 = a_1 + 2d, \quad \ldots, \quad a_n = a_1 + (n-1)d \]

2. **Rewrite the sum**:
We need to evaluate the sum:
\[ S = \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \ldots + \frac{1}{a_{n-1} a_n} \]
This can be rewritten using the terms of the A.P.:
\[ S = \frac{1}{a_1(a_1 + d)} + \frac{1}{(a_1 + d)(a_1 + 2d)} + \ldots + \frac{1}{(a_1 + (n-2)d)(a_1 + (n-1)d)} \]

3. **Use the identity for fractions**:
Notice that:
\[ \frac{1}{a_k a_{k+1}} = \frac{1}{d} \left( \frac{1}{a_k} - \frac{1}{a_{k+1}} \right) \]
This means we can express each term in the sum \( S \) as:
\[ S = \frac{1}{d} \left( \left( \frac{1}{a_1} - \frac{1}{a_2} \right) + \left( \frac{1}{a_2} - \frac{1}{a_3} \right) + \ldots + \left( \frac{1}{a_{n-1}} - \frac{1}{a_n} \right) \right) \]

4. **Simplify the sum**:
The terms in the parentheses form a telescoping series:
\[ S = \frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_n} \right) \]

5. **Express \( a_n \)**:
From the A.P. formula, we have:
\[ a_n = a_1 + (n-1)d \quad \Rightarrow \quad d = \frac{a_n - a_1}{n-1} \]

6. **Substitute \( d \) back into the sum**:
Now substitute \( d \) into the expression for \( S \):
\[ S = \frac{1}{\frac{a_n - a_1}{n-1}} \left( \frac{1}{a_1} - \frac{1}{a_n} \right) = \frac{n-1}{a_n - a_1} \left( \frac{1}{a_1} - \frac{1}{a_n} \right) \]

7. **Combine the fractions**:
Simplifying the expression gives:
\[ S = \frac{n-1}{(a_n - a_1) a_1 a_n} \]

8. **Final result**:
Since \( a_n - a_1 = (n-1)d \), we can express \( S \) as:
\[ S = \frac{n-1}{a_1 a_n} \]

Thus, we have proved that:

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