Problem 71 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(b\) is the first term of an infinite G. P whose sum is five, then \(b\) lies in the interval. \(\quad\) [Online April 15, 2018] (a) \((-\infty,-10)\) (b) \((10, \infty)\) (c) \((0,10)\) (d) \((-10,0)\)

Step-by-Step Solution

Answer

The interval is (0,10).

Step 1: Understand the formula for the sum of an infinite geometric series

The sum of an infinite geometric series with first term \(a\) and common ratio \(r\) (where \(|r| < 1\)) is given by the formula: \[ S = \frac{a}{1-r} \] This is key to solving the problem because we know the sum, and we can use this formula to express the relationship between terms.

Step 2: Identify the known values and expressions

We are told that \(b\) is the first term of the geometric series, and the sum \(S = 5\). We can substitute these values into the formula: \[ 5 = \frac{b}{1-r} \] This equation will help us find the acceptable range of \(b\).

Step 3: Solve for common ratio \(r\) in terms of \(b\)

Rearrange the formula to find \(r\): \[ 1 - r = \frac{b}{5} \] Thus, \[ r = 1 - \frac{b}{5} \] Since the series is infinite, \(|r| < 1\). This condition will help set boundaries for \(b\).

Step 4: Apply the condition \(|r| < 1\) to find bounds for \(b\)

Substitute \(r\) from Step 3 into \(|r| < 1\): \[ \left|1 - \frac{b}{5}\right| < 1 \] Consider the two inequality conditions: 1. \(1 - \frac{b}{5} < 1\) 2. \(1 - \frac{b}{5} > -1\)

Step 5: Solve the inequalities

Solve the inequalities separately: 1. \(1 - \frac{b}{5} < 1\) This simplifies to: \(\frac{b}{5} > 0\), or \(b > 0\)2. \(1 - \frac{b}{5} > -1\) This simplifies to: \(b < 10\)Combining these, \(0 < b < 10\).

Step 6: Identify the correct interval from options

The solution from Step 5, \(0 < b < 10\), corresponds to option (c). Therefore, the interval where \(b\) lies is \((0,10)\).

Key Concepts

Sum of Infinite Series

An infinite geometric series is a sum of terms that continues indefinitely. The series has the form \( a, ar, ar^2, ar^3, \ldots \) and so on. To find the sum of these infinite terms, we use a special formula:

• If the absolute value of the common ratio \( r \) is less than 1 (\( |r| < 1 \)), the series converges. This means it approaches a fixed sum.
• The sum \( S \) of an infinite geometric series with the first term \( a \) is given by \( S = \frac{a}{1-r} \).

This formula allows us to find a fixed sum, even though there are infinitely many terms. It's crucial for solving problems involving infinite series, like determining the possible values for the first term in specific scenarios, such as our exercise.

Geometric Progression

A geometric progression (G.P.) is an important concept where each term after the first is found by multiplying the previous term by a constant, known as the common ratio \( r \). In a G.P., if the first term is \( a \) and the common ratio is \( r \), the series looks like this:

• The first term: \( a \)
• The second term: \( ar \)
• The third term: \( ar^2 \)
• And so on: \( ar^3, ar^4, \ldots \)

Understanding the structure of a geometric progression helps us see how terms relate to one another. It is the backbone of forming equations for both finite series (where the progression ends) and infinite series (where the pattern continues indefinitely). In our exercise, \( b \) represents the first term of such a progression.

Common Ratio Condition

The common ratio \( r \) of a geometric series plays a key role in determining whether the series is convergent (approaches a sum) or divergent (grows indefinitely). For an infinite geometric series to converge, the common ratio must satisfy the condition \( |r| < 1 \).So what does this condition mean?

• \(|r| < 1\) ensures that as we multiply the first term by \( r \) repeatedly, the terms get smaller and eventually become negligible.
• This shrinking of terms allows the infinite series to approach a finite limit or sum.
• In our exercise, ensuring \( |r| < 1 \) helps establish the interval that confines the value of the first term \( b \).

Applying this condition ensures that the series does not spiral into infinity, which is key to calculating the sum of the series reliably.