Problem 82 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If the sum of \(m\) terms of an A.P. is equal to sum of either the next \(n\) terms or the next \(p\) terms, prove that \((m+n)\left(\frac{1}{m}-\frac{1}{p}\right)=(m+p)\left(\frac{1}{m}-\frac{1}{n}\right) .\)

Step-by-Step Solution

Answer

Given that the sum of $m$ terms of an A.P. equals the sum of either the next $n$ terms or the next $p$ terms, we proved that \((m+n)\left(\frac{1}{m}-\frac{1}{p}\right) = (m+p)\left(\frac{1}{m}-\frac{1}{n}\right)\) by forming the arithmetic progression, substituting the general formulae for the sums, simplifying, and canceling out common factors.

Step 1: Form the arithmetic progression and given condition

Let the A.P. be given by \(a, a+d, a+2d, ...\). According to the problem, the sum of first 'm' terms equals the sum of either the next 'n' terms or the next 'p' terms. So, we have: \[S_m = S_{m+n} - S_m = S_{m+p} - S_m\]

Step 2: Write general formulae for the sums

As we know, the sum of first 'k' terms of an A.P is given by S_k = \(\frac{k}{2}(2a + (k-1)d)\). Using this formula, we can write the sums for m, m+n and m+p terms. \[S_m = \frac{m}{2}(2a + (m-1)d)\] \[S_{m+n} = \frac{m+n}{2}(2a + (m+n-1)d)\] \[S_{m+p} = \frac{m+p}{2}(2a + (m+p-1)d)\]

Step 3: Substitute and simplify

Now we will substitute the given sums in the main equation and simplify. \[\frac{m+n}{2}(2a + (m+n-1)d) - \frac{m}{2}(2a + (m-1)d) = \frac{m+p}{2}(2a + (m+p-1)d) - \frac{m}{2}(2a + (m-1)d)\] Cancel out the common terms on both sides and simplify further: \[(m+n)\left(\frac{1}{m}-\frac{1}{p}\right)d = (m+p)\left(\frac{1}{m}-\frac{1}{n}\right)d\]

Step 4: Cancel out common factors

Now, we can cancel out the 'd' term from both sides to get our final proof: \[(m+n)\left(\frac{1}{m}-\frac{1}{p}\right) = (m+p)\left(\frac{1}{m}-\frac{1}{n}\right)\] Hence, we have proved that if the sum of 'm' terms of an A.P. is equal to the sum of either the next 'n' terms or the next 'p' terms, then \((m+n) \left(\frac{1}{m}-\frac{1}{p}\right)=(m+p) \left(\frac{1}{m}-\frac{1}{n}\right)\).

Key Concepts

Sum of Arithmetic Progression

An arithmetic progression (A.P.) is a sequence of numbers in which the difference of any two successive members is a constant. This constant is known as the "common difference". The sequence can be written as: \( a, a+d, a+2d, ... \), where \( a \) is the first term and \( d \) is the common difference.
The sum of the first 'k' terms of an arithmetic progression can be calculated using the formula:

• \( S_k = \frac{k}{2}(2a + (k-1)d) \)

This formula is derived by adding the first and last term and multiplying by the number of pairs, similar to finding the area of a rectangle. This helps simplify the process of summing up a large number of terms in sequences like arithmetic progressions.
In the given problem, understanding how to express the sum of 'm', 'm+n', and 'm+p' terms using this formula is crucial in setting up the equation for proving the relationship between these sums.

Proof in Algebra

Algebraic proof involves demonstrating the truth of an equation or statement using algebraic techniques. In proving equalities or identities in algebra, manipulation of expressions using known formulae and identities is key. This approach requires logical reasoning and methodical work through each step to ensure validity.
In the original problem, the algebraic proof uses manipulation of the sums of arithmetic series. By writing the sums for \(S_m\), \(S_{m+n}\), and \(S_{m+p}\) using the known formula for arithmetic sequences, we can set the two conditions equal based on the problem's description.

• You begin by equating \(S_{m+n} - S_m \) with \( S_{m+p} - S_m \).
• This allows for canceling out of terms and simplifying the remaining expression.

In the end, with careful simplification and canceling of common factors, you reach the final equation that needs to be proved. Every step must ensure logical consistency to establish the equation's correctness.

Mathematical Induction

Mathematical induction is a powerful proof technique used particularly for proving statements about integers. It consists of two major steps: the base case and the inductive step. Although this technique is not directly applied in the given problem, understanding induction can be useful for similar proofs where sequences and series are involved.

• Base Case: Verify that the initial case of the statement is true. For instance, proving for \(n = 1\).
• Inductive Step: Assume the statement holds for \(n = k\), and then prove it for \(n = k+1\).

Using induction often simplifies proofs by establishing a pattern or rule that holds across all integers. When done correctly, it confirms that the statement is true for all subsequent values beyond the base case.
While induction wasn't applied here, understanding it can broaden your proof techniques toolkit for similar mathematical problems.