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A sequence is simply an ordered list of numbers. In an arithmetic progression (AP), each term after the first is obtained by adding a fixed number, known as the common difference, to the previous term. This yields a set of terms that increase or decrease consistently throughout the sequence. In the exercise, the first three terms are provided as: \( \frac{1}{1+\sqrt{x}} \), \( \frac{1}{1-x} \), and \( \frac{1}{1-\sqrt{x}} \). To ascertain if these terms form an arithmetic sequence, we must verify that the difference between consecutive terms remains constant throughout. Once the difference is verified, knowing these terms allows us to explore the sequence further, like finding additional terms or deducing the general pattern they follow.

The common difference is the specific number that separates each pair of consecutive terms in an arithmetic progression. Calculating this value is key to confirming whether a sequence is indeed an AP.To calculate the common difference for the given sequence:- Let the terms be \( a = \frac{1}{1+\sqrt{x}} \), \( b = \frac{1}{1-x} \), and \( c = \frac{1}{1-\sqrt{x}} \).- Calculating the difference between the first two terms (\(b - a\)), results in: \[ b - a = \frac{1}{1-x} - \frac{1}{1+\sqrt{x}} = \frac{x+\sqrt{x}}{(1-x)(1+\sqrt{x})} \]- Similarly, find the difference between the second and third terms (\(c - b\)): \[ c - b = \frac{1}{1-\sqrt{x}} - \frac{1}{1-x} = \frac{x-\sqrt{x}}{(1-\sqrt{x})(1-x)} \]Both differences simplify to equivalent expressions, thereby affirming that the sequence is indeed an arithmetic progression.

The n-th term of an arithmetic progression can be found using a straightforward formula. Once you have the first term and the common difference, you can compute any term in the sequence with ease. The n-th term formula for an AP is: \[ T_n = a + (n - 1) \cdot d \]Where:

• \( T_n \) is the n-th term we want to find,
• \( a \) is the first term,
• \( d \) is the common difference,
• \( n \) is the position of the term in the sequence.

For the given sequence, substitute the calculated values:- First term, \( a = \frac{1}{1+\sqrt{x}} \)- Common difference, \( d = \frac{x+\sqrt{x}}{(1-x)(1+\sqrt{x})} \)Thus, the n-th term is given by:\[ T_n = \frac{1}{1+\sqrt{x}} + (n-1) \cdot \frac{x+\sqrt{x}}{(1-x)(1+\sqrt{x})} \]This formula enables quick computation of any term in the sequence, providing a comprehensive understanding of its pattern.