Problem 67 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is \(\frac{27}{19}\). Then the common ratio of this series is: (a) \(\frac{1}{3}\) (b) \(\frac{2}{3}\) (c) \(\frac{2}{9}\) (d) \(\frac{4}{9}\)

Step-by-Step Solution

Answer

The common ratio is \( \frac{2}{3} \).

Step 1: Identify the Formula for the Sum

For an infinite geometric series with first term \(a\) and common ratio \(r\), the sum is \( \frac{a}{1-r} \) where \(|r| < 1\). We are given that this sum is 3, so we have the equation \( \frac{a}{1-r} = 3 \).

Step 2: Use the Sum of Cubes Formula

The sum of the cubes of an infinite geometric series is \( \frac{a^3}{1-r^3} \), which is given to be \( \frac{27}{19} \). Thus, the equation is \( \frac{a^3}{1-r^3} = \frac{27}{19} \).

Step 3: Relate the Two Equations

From Step 1, we have \( a = 3(1-r) \). Substitute \(a\) in Step 2's equation: \( \frac{(3(1-r))^3}{1-r^3} = \frac{27}{19} \).

Step 4: Simplify and Solve for r

Simplifying, we get \( \frac{27(1-r)^3}{1-r^3} = \frac{27}{19} \). Cancel 27 from both sides to get \( \frac{(1-r)^3}{1-r^3} = \frac{1}{19} \).

Step 5: Solve the Simplified Equation

Expand \((1-r)^3\) to get \(1 - 3r + 3r^2 - r^3\). The equation becomes \( \frac{1 - 3r + 3r^2 - r^3}{1-r^3} = \frac{1}{19} \). Simplifying further yields \(1 - 3r + 3r^2 = \frac{1}{19}(1 - r^3)\).

Step 6: Substitute Possible Values

Substitute the possible values of \(r\) from the options into the equation derived in Step 5 to find which value satisfies it. After checking, \(r = \frac{2}{3}\) satisfies the equation.

Key Concepts

Sum of an Infinite Geometric Series

In an infinite geometric series, terms continue indefinitely and follow a specific pattern whereby each term is a fixed fraction of its preceding term. This fixed fraction is known as the common ratio. The sum of an infinite geometric series converges, which means it approaches a finite value, only when the absolute value of the common ratio is less than 1, or \(|r| < 1\).
For such a series, the sum is calculated through the formula:

• \(S = \frac{a}{1 - r}\)

where:

• \(S\) is the sum of the series,
• \(a\) is the first term of the series, and
• \(r\) is the common ratio.

In the given problem, the sum is given as 3, indicating that:

• \(\frac{a}{1 - r} = 3\)

By understanding this principle, you can determine the relationship between the first term and the common ratio, which helps solve various problems involving geometric series.

Sum of Cubes of Geometric Series Terms

The sum of the cubes of the terms in a geometric series involves a slightly more complex concept. When each term of the series is cubed, it forms a new series. For an infinite series with first term \(a\) and common ratio \(r\), this newly formed series can also converge and its sum is represented as:

• \(\text{Sum of cubes} = \frac{a^3}{1 - r^3}\)

In the exercise, the sum of the cubes is given as \(\frac{27}{19}\), leading to the equation:

• \(\frac{a^3}{1 - r^3} = \frac{27}{19}\)

The challenge is to find \(r\) while making sure both the basic sum and the sum of cubes hold true. You need to express these terms in relation to each other and solve. Understanding the sum of cubes allows you to analyze more complex patterns high school students encounter frequently in advanced math courses.

Common Ratio

The common ratio \(r\) is a crucial element in defining a geometric series. It is the ratio between each consecutive term and can greatly impact both the convergence and the sum of the series.
The common ratio is calculated simply as:

• For a sequence \(a, ar, ar^2, ar^3, \ldots\), \(r = \frac{\text{next term}}{\text{previous term}}\)

In exercises like the one provided, examining different potential common ratios is key. Values like \(\frac{1}{3}, \frac{2}{3}, \frac{2}{9},\) and \(\frac{4}{9}\) reflect options that need testing in specific equations obtained from given conditions.
The common ratio affects:

• The speed at which the series converges,
• The size of the sum of an infinite series, and
• The formation of related new series (like cubes of terms).

Identifying the right common ratio requires substituting these values into the equations derived from series properties, ensuring that the appropriate value validates all given conditions. This task deepens your understanding of how geometric series operate and empowers problem-solving with series.