Problem 12 - A Complete Resource Book in Mathematics for JEE Main

Question

The sum of the first hundred terms of an A.P. is \(x\) and the sum of the hundred terms starting from the third term is \(y\). Then the common difference is (A) \(\frac{y-x}{2}\) (B) \(\frac{y-x}{50}\) (C) \(\frac{y-x}{100}\) (D) \(\frac{y-x}{200}\)

Step-by-Step Solution

Answer

The common difference is (D) \(\frac{y-x}{200}\).

Step 1: Understand the Problem

An arithmetic progression (A.P.) is a sequence of numbers where the difference between consecutive terms is constant. We are given that the sum of the first 100 terms is \(x\) and the sum of 100 terms starting from the third term is \(y\). We need to find the common difference of the A.P.

Step 2: Express the Sums in A.P. Form

The sum of the first \(n\) terms of an A.P. is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\), where \(a\) is the first term and \(d\) is the common difference. For the first 100 terms:\[ S_{100} = \frac{100}{2} (2a + 99d) = 50(2a + 99d) = x \]

Step 3: Express the Second Sum in A.P. Form

The sum of the 100 terms starting from the third term is:\[ S_{102} - S_2 = \frac{100}{2} (2(a + 2d) + 99d) = 50(2a + 4d + 99d) = 50(2a + 103d) = y \]

Step 4: Form Equations and Simplify

From Steps 2 and 3, we have two equations:1. \(50(2a + 99d) = x\)2. \(50(2a + 103d) = y\)Subtract Equation 1 from Equation 2 to find:\[ 50(2a + 103d) - 50(2a + 99d) = y - x \]This simplifies to:\[ 50(4d) = y - x \]

Step 5: Solve for the Common Difference

From the simplified equation \(50(4d) = y - x\), solve for \(d\):\[ 200d = y - x \]\[ d = \frac{y - x}{200} \]

Step 6: Determine the Correct Option

The formula for the common difference we found is \(d = \frac{y-x}{200}\). Comparing with the given options, option (D) \(\frac{y-x}{200}\) correctly matches our answer.

Key Concepts

Sum of n terms

In an arithmetic progression (A.P.), the sum of the first n terms is a fundamental concept. To find this sum, we use the formula: \[ S_n = \frac{n}{2} (2a + (n-1)d) \] where:

• \(S_n\) is the sum of the first n terms,
• \(a\) is the first term, and
• \(d\) is the common difference.

This formula essentially captures how many terms you're adding, \((n/2)\), times the average of the first and last terms. For example, if you wanted to find the sum of the first 100 terms, you would substitute n with 100.
In many problems, knowing how to appropriately set up your terms is as crucial as calculating the sum. When given certain conditions, you often solve for unknowns like the first term or the common difference by using two or more instances of \(S_n\). Understanding this approach will give you a solid foundation in handling similar problems across A.P. series.

Common Difference

The common difference in an arithmetic progression is at the heart of how sequences behave. Represented by \(d\), it defines the difference between consecutive terms in the sequence.
To find the common difference, one often relies on previously calculated sums or known values within the sequence. As demonstrated in the solution, starting from deriving two equations for different parts of a sequence enabled solving for \(d\).
By subtracting the sum equation of the first part from the sum of the subsequent part and simplifying, you isolate the portion that focuses on the common difference. This technique of manipulating the sums aids in solving \( 200d = y - x \)for \(d\), arriving at the formula:\[ d = \frac{y - x}{200} \]Understanding this calculation can help reinforce how changes in one part of a sequence trigger shifts throughout it.

Sequences and Series

In mathematics, sequences and series are closely related concepts, especially in arithmetic progressions. A sequence is simply a list of numbers following a rule, like an arithmetic sequence where each number is a certain common difference away from the one before it. Points in a sequence are often denoted by terms like \(a, a+d, a+2d,\) etc.
A series, meanwhile, is the sum of the terms in a sequence. In an arithmetic series, adding a common difference repeatedly results in lines that grow at a constant rate.
The understanding of how sequences turn into series is key to solving problems related to arithmetic progressions. This encompasses applying formulas, managing given conditions, and using logical deduction to connect different parts of the progression. As you continue to work with A.P. problems, treat sequences as the step-by-step roadmap and the series as the overarching journey they sum up to.