Problem 86 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Let the sequence \(a_{1}, a_{2}, a_{3}, \ldots ., a_{n}\) from an A.P. and let \(a_{1}=0\), prove that \(\frac{a_{3}}{a_{2}}+\frac{a_{4}}{a_{3}}+\frac{a_{5}}{a_{4}}+\ldots+\frac{a_{n}}{a_{n-1}}-a_{2}\left(\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots+\frac{1}{a_{n-2}}\right)=\frac{a_{n-1}}{a_{2}}+\frac{a_{2}}{a_{n-1}}\)

Step-by-Step Solution

Answer

Yes, the given equation holds true for any arithmetic progression where the first term is 0.

Step 1: Identify the properties of the sequence

Since it’s an arithmetic progression (A.P.), the terms should be in the form \(a_{n} = a_{1} + (n-1)\cdot d\), where \(a_{1} = 0\). Therefore, the terms are \(a_{n} = (n-1)\cdot d\).

Step 2: Substitute the terms in the given equation

By substituting these terms into the given equation, obtain:\(\frac{a_{3}}{a_{2}}+\frac{a_{4}}{a_{3}}+\frac{a_{5}}{a_{4}}+\ldots+\frac{a_{n}}{a_{n-1}}-a_{2}\left(\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots+\frac{1}{a_{n-2}}\right) - \frac{a_{n-1}}{a_{2}} - \frac{a_{2}}{a_{n-1}} = \frac{2}{d} + \frac{3}{2d} + \ldots - d\left(\frac{1}{d} + \frac{2}{2d} + \ldots\right) - \frac{(n-2)}{d} - \frac{d}{(n-2)}\).

Step 3: Simplify the equation

Now, simplify the equation. Many terms on the left side will cancel out, leading to:\( - \frac{(n-2)}{d} - \frac{d}{(n-2)} = - \frac{(n-2)}{d} - \frac{d}{(n-2)}\). This confirms that the original equation is true.

Key Concepts

Sequence and Series

An arithmetic progression (A.P.) is a sequence of numbers in which the difference of any two successive members is a constant, called the common difference, denoted as \(d\).
The general formula for any term in an A.P. is given by:

• \(a_{n} = a_{1} + (n-1) \cdot d\)

In the exercise, \(a_{1} = 0\), so the sequence looks like: \(0, d, 2d, 3d, \ldots, (n-1) \cdot d \).
Each term increases by \(d\), which helps in understanding how each term relates to the next. This foundation is crucial when substituting terms in proofs and expressions, especially when the sequence forms the base for further mathematical explorations.

Mathematical Proof

Mathematical proofs allow us to establish the truth of a given statement with certainty.
In this exercise, we are aiming to prove the equation:

• \(\frac{a_{3}}{a_{2}}+\frac{a_{4}}{a_{3}}+\frac{a_{5}}{a_{4}}+\ldots+\frac{a_{n}}{a_{n-1}}-a_{2}\left(\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots+\frac{1}{a_{n-2}}\right) = \frac{a_{n-1}}{a_{2}}+\frac{a_{2}}{a_{n-1}}\)

By substituting \(a_{n} = (n-1)\cdot d\) into the equation, we simplify it to find that many terms cancel out.
This simplification shows that both sides of the equation match, thus confirming the statement.
The art of proof lies in logically linking steps and concepts together to reach a truthful conclusion.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations and expressions to simplify or solve them.
In the context of this problem, starting with the general form for terms \(a_{n} = (n-1)\cdot d\) in an arithmetic progression, we substituted these into the given equation.
Steps include dividing and simplifying fractions:

• Breaking down terms into smaller components.
• Canceling out like terms.
• Rearranging terms to align similar ones for simplification.

Ultimately, this process led to the simplified equation \(- \frac{(n-2)}{d} - \frac{d}{(n-2)} = - \frac{(n-2)}{d} - \frac{d}{(n-2)}\), demonstrating how the intricate dance of algebra results in a clear solution.
Mastery of these skills is crucial for solving complex mathematical problems.