Problem 156 - A Complete Resource Book in Mathematics for JEE Main

Question

If 100 times the \(100^{\text {th }}\) term of an Arithmetic Progression with non zero common difference equals the 50 times its \(50^{\text {th }}\) term, then the \(150^{\text {th }}\) term of this A.P. is |2012| (A) \(-150\) (B) 150 times its \(50^{\text {th }}\) term (C) 150 (D) zero

Step-by-Step Solution

Answer

The 150th term of the AP is zero.

Step 1: Understanding the Problem

We need to compare the terms of an arithmetic progression (AP) at specific positions. We're given that 100 times the \(100^{\text{th}}\) term equals 50 times the \(50^{\text{th}}\) term. We want to find the \(150^{\text{th}}\) term.

Step 2: Formulating the General Term

The general term \(T_n\) of an AP can be written as \(a + (n-1)d\) where \(a\) is the first term and \(d\) is the common difference.

Step 3: Expressing Specific Terms

Using the general formula, the \(100^{\text{th}}\) term is \(T_{100} = a + 99d\) and the \(50^{\text{th}}\) term is \(T_{50} = a + 49d\).

Step 4: Setting Up the Given Equation

According to the problem, we have \(100 \, T_{100} = 50 \, T_{50}\). Substituting the expressions for these terms gives us: \[100(a + 99d) = 50(a + 49d).\]

Step 5: Simplifying the Equation

Expanding and simplifying the equation: \[100a + 9900d = 50a + 2450d.\] This simplifies to \[50a + 7450d = 0.\] Hence, \[a = -149d.\]

Step 6: Finding the 150th Term

The \(150^{\text{th}}\) term is \(T_{150} = a + 149d \). Substituting \(a = -149d\) gives \[T_{150} = -149d + 149d = 0.\] This means the \(150^{\text{th}}\) term is zero.

Key Concepts

Common Difference

In an Arithmetic Progression (AP), the term "Common Difference" refers to the constant amount that each term in the sequence increases or decreases by as you move from one term to the next. This is a defining feature of an AP, making it distinct from other sequences.

The common difference, denoted as \(d\), can be positive, negative, or zero, although the given exercise specifies a non-zero common difference. To find the common difference in a specific AP, you can subtract any term from the term that follows it. For example, if you have the sequence 2, 5, 8, 11, the common difference \(d\) would be 3 because 5 - 2 = 3. Similarly, 8 - 5 = 3.

Understanding the common difference is crucial because it helps in exploring the behavior and pattern within an AP. It's used to formulate the general term and calculate specific terms in the series.

General Term of AP

The general term of an Arithmetic Progression (AP) is very important as it provides a formula to find any term in the sequence without needing to list all the previous terms. The general term is expressed as \(T_n = a + (n - 1)d\), where \(a\) represents the first term, \(d\) is the common difference, and \(n\) is the term number you want to find.

This formula is very efficient because:

• It provides a straightforward way to find any term in the AP
• It helps in setting up equations like in the given problem, where terms at specific positions are compared

To grasp this better, let's consider an AP with the first term \(a = 2\) and a common difference \(d = 3\). The third term \(T_3\) would be calculated as: \(T_3 = 2 + (3-1) \, 3 = 8\).

By using the formula for the general term, you can solve more complex problems involving arithmetic progressions without confusion.

Solving Equations

When working with arithmetic progressions, solving equations is a common task, especially when we have relationships between terms at different positions. In the context of the exercise, we had a scenario where certain terms multiplied by numbers equaled each other. This leads to the equation \(100(a + 99d) = 50(a + 49d)\).

To solve such equations, you typically:

• Substitute expressions of terms using the general term formula
• Expand and simplify the equation
• Solve for unknowns such as \(a\) or \(d\)

In our exercise, simplifying the equation gives us \(50a + 7450d = 0\). Solving further, we find \(a = -149d\).

These steps help reduce complex problems into simpler arithmetic operations, allowing us to find specific terms efficiently. Once the equation is solved, you can find any term in the sequence, like the \(150^{\text{th}}\) term, by substituting the known values of \(a\) and \(d\), which in this case resulted in a zero value for the term.