Problem 20 - A Complete Resource Book in Mathematics for JEE Main

Question

The G.M. of the number \(3,3^{2}, 3^{3}, \ldots, 3^{3 n}\) is (A) \(3^{\frac{n}{2}}\) (B) \(3^{\frac{3 n}{2}}\) (C) \(3^{\frac{3 n+1}{2}}\) (D) \(3^{\frac{n+1}{2}}\)

Step-by-Step Solution

Answer

The G.M. is \(3^{\frac{3n+1}{2}}\). Option (C) is correct.

Step 1: Identify the Terms

The sequence given is a geometric progression: \(3, 3^2, 3^3, \ldots, 3^{3n}\). The number of terms in this sequence is \(3n\) because it starts from \(3^1\) and goes to \(3^{3n}\).

Step 2: Formula for Geometric Mean

To find the geometric mean (G.M.) of a series of numbers, use the formula: \((a_1 \cdot a_2 \cdot a_3 \cdots a_n)^{1/n}\). Here, \(n\) is the number of terms and \(a_1, a_2, \ldots, a_n\) are the terms.

Step 3: Calculate the Product of the Terms

The product of all terms in the sequence can be expressed as: \(3 \times 3^2 \times 3^3 \times \ldots \times 3^{3n}\). This simplifies to \(3^{1+2+3+\ldots+3n}\).

Step 4: Use Sum of an Arithmetic Sequence

The sum of the exponents \(1+2+3+\ldots+3n\) is given by the formula for the sum of an arithmetic series: \(S = \frac{n}{2} \cdot (a+l)\), where \(a=1\) and \(l=3n\). Hence, \(S = \frac{3n(1+3n)}{2} = \frac{3n(3n+1)}{2}\).

Step 5: Calculate the Geometric Mean

Now substitute into the geometric mean formula: \((3^{\frac{3n(3n+1)}{2}})^{\frac{1}{3n}} = 3^{\frac{3n+1}{2}}\).

Step 6: Match with the Given Options

From the calculated geometric mean \(3^{\frac{3n+1}{2}}\), we see that it matches option (C).

Key Concepts

Geometric Progression

Let's begin by exploring the concept of a geometric progression. A geometric progression, or GP, is a sequence in which each term is found by multiplying the previous term by a constant known as the "common ratio." In the sequence given, which is composed of powers of 3: \(3, 3^2, 3^3, \ldots, 3^{3n}\), the common ratio is 3.
This means that each term is derived by multiplying the previous term by 3, making it a perfect example of a geometric progression.

• First term: 3
• Secondary term: \(3^2\)
• Subsequent terms continue multiplying by 3

The number of terms is determined by the highest power of 3 in the sequence, which in this case is \(3^{3n}\). Therefore, there are \(3n\) terms. Understanding this progression helps when calculating items like the geometric mean, as it reveals the uniform pattern shared by all terms.

Sum of Arithmetic Sequence

A key step in solving for the geometric mean involves evaluating the sum of the exponents in the sequence. While the original sequence is a geometric progression, the pattern in the exponents \(1, 2, 3, \ldots, 3n\) forms an arithmetic sequence. An arithmetic sequence is characterized by a constant difference between consecutive terms. Here, the difference is 1.

To find the sum of this arithmetic sequence, use the formula:\[S = \frac{n}{2} \cdot (a + l),\]where:

• \(n\) is the number of terms
• \(a\) is the first term
• \(l\) is the last term

For our sequence:

• First term \(a = 1\)
• Last term \(l = 3n\)
• Number of terms \(n = 3n\)

The sum is calculated as \(S = \frac{3n(1 + 3n)}{2} = \frac{3n(3n + 1)}{2}\). This sum of the exponents is critical for determining the geometric mean of the sequence.

Mathematical Sequence

Mathematical sequences are ordered lists of numbers that have a specific relationship between their terms. The sequence type helps define the relationships and methods for analysis, like calculating means or sums.

In our example, we first recognize that the base sequence is geometric. However, examining the exponents presents an arithmetic sequence, reflective of mathematics' interconnected nature. Each kind serves different purposes.

By understanding both sequences:

• Geometric Progression (GP) informs about the multiplicative relationship between terms.
• Arithmetic Sequence underlies the additive pattern of exponents.

Utilizing the properties of these sequences enables complex calculations, such as deriving the geometric mean from the given sequence. Being equipped with knowledge of both concepts helps solve exercises where these overlap, reinforcing comprehensive mathematical understanding.