Let the positive numbers $a_1, a_2, a_3, ...., a_{3n}$ be in GP. If P is the GM of $a_1, a_2, a_3, ...., a_n$ and Q is the GM of $a_{n+1}, a_{n+2}, a_{n+3}, ...., a_{3n}$, then what is the GM of $3n$ numbers?

Question

Prepp Team · Accepted Answer

Understanding Geometric Progression and Mean

A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The terms of a GP can be represented as $a, ar, ar^2, \dots$, where $a$ is the first term.

The Geometric Mean (GM) of a set of $k$ non-negative numbers $x_1, x_2, \dots, x_k$ is calculated using the formula:

$ GM = (x_1 \cdot x_2 \cdot ... \cdot x_k)^{\frac{1}{k}} $

In this problem, we are given $3n$ positive numbers ($a_1, a_2, \dots, a_{3n}$) that form a GP. We are provided with P, the GM of the first $n$ terms ($a_1, \dots, a_n$), and Q, the GM of the subsequent $2n$ terms ($a_{n+1}, \dots, a_{3n}$). The objective is to determine the GM of all $3n$ numbers.

Setting up the Geometric Progression

Let the GP sequence be denoted by $a_k$, where $k = 1, 2, \dots, 3n$. We can express the terms using the first term $a$ and the common ratio $r$. Since the numbers are positive, we know $a > 0$ and $r > 0$. The terms are:

$ a_k = a \cdot r^{k-1} $

So the sequence is $a, ar, ar^2, \dots, ar^{3n-1}$.

Calculating P: GM of the First $n$ Terms

P is given as the Geometric Mean of the first $n$ terms: $a_1, a_2, \dots, a_n$. These terms are $a, ar, ar^2, \dots, ar^{n-1}$.

Applying the GM formula for these $n$ terms:

$ P = (a_1 \cdot a_2 \cdot ... \cdot a_n)^{\frac{1}{n}} $

$ P = \left( a \cdot (ar) \cdot (ar^2) \cdot ... \cdot (ar^{n-1}) \right)^{\frac{1}{n}} $

Group the $a$ terms and the $r$ terms:

$ P = \left( a^n \cdot r^{0+1+2+...+(n-1)} \right)^{\frac{1}{n}} $

The sum of the exponents of $r$ is an arithmetic series sum: $0+1+2+...+(n-1) = \frac{n(n-1)}{2}$.

Substituting the sum back:

$ P = \left( a^n \cdot r^{\frac{n(n-1)}{2}} \right)^{\frac{1}{n}} $

Simplify by distributing the exponent $\frac{1}{n}$:

$ P = a^{\frac{n}{n}} \cdot r^{\frac{n(n-1)}{2n}} = a^1 \cdot r^{\frac{n-1}{2}} $

So, $P = a \cdot r^{\frac{n-1}{2}}$.

Calculating Q: GM of Terms $a_{n+1}$ to $a_{3n}$

Q is the Geometric Mean of the terms $a_{n+1}, a_{n+2}, \dots, a_{3n}$. These terms correspond to $ar^n, ar^{n+1}, \dots, ar^{3n-1}$.

The number of terms in this group is $(3n) - (n+1) + 1 = 3n - n - 1 + 1 = 2n$.

Applying the GM formula for these $2n$ terms:

$ Q = (a_{n+1} \cdot a_{n+2} \cdot ... \cdot a_{3n})^{\frac{1}{2n}} $

$ Q = \left( (ar^n) \cdot (ar^{n+1}) \cdot ... \cdot (ar^{3n-1}) \right)^{\frac{1}{2n}} $

Group the terms:

$ Q = \left( a^{2n} \cdot r^{n + (n+1) + ... + (3n-1)} \right)^{\frac{1}{2n}} $

The sum of the exponents $n + (n+1) + \dots + (3n-1)$ is an arithmetic series sum. The number of terms is $2n$. The sum is $\frac{2n}{2} \times (\text{first term} + \text{last term}) = n \times (n + (3n-1)) = n(4n-1) = 4n^2 - n$.

Substituting the sum back:

$ Q = \left( a^{2n} \cdot r^{4n^2 - n} \right)^{\frac{1}{2n}} $

Simplify by distributing the exponent $\frac{1}{2n}$:

$ Q = a^{\frac{2n}{2n}} \cdot r^{\frac{4n^2 - n}{2n}} = a^1 \cdot r^{\frac{n(4n-1)}{2n}} = a \cdot r^{\frac{4n-1}{2}} $

So, $Q = a \cdot r^{\frac{4n-1}{2}}$.

Evaluating Option 3: $\sqrt{PQ}$

We need to find the GM of all $3n$ numbers. Let's evaluate the expression given in Option 3, $\sqrt{PQ}$, using the derived formulas for P and Q.

$ \sqrt{PQ} = \sqrt{\left( a \cdot r^{\frac{n-1}{2}} \right) \cdot \left( a \cdot r^{\frac{4n-1}{2}} \right)} $

Multiply the terms inside the square root:

$ \sqrt{PQ} = \sqrt{a^2 \cdot r^{\left(\frac{n-1}{2} + \frac{4n-1}{2}\right)}} $

First, simplify the exponent of $r$:

$ \frac{n-1}{2} + \frac{4n-1}{2} = \frac{n-1 + 4n-1}{2} = \frac{5n-2}{2} $

Now substitute this back into the expression:

$ \sqrt{PQ} = \sqrt{a^2 \cdot r^{\frac{5n-2}{2}}} $

Finally, take the square root:

$ \sqrt{PQ} = a \cdot r^{\frac{5n-2}{4}} $

This calculation shows the value of the expression $\sqrt{PQ}$.

Conclusion

The problem asks for the Geometric Mean of the $3n$ numbers in the GP. We have calculated the values of P and Q based on their definitions. By evaluating Option 3, $\sqrt{PQ}$, we obtained the expression $a \cdot r^{\frac{5n-2}{4}}$. This represents one of the potential forms for the overall GM of the sequence.

Based on the provided options and the calculations performed, the correct representation for the Geometric Mean of the $3n$ numbers is $\sqrt{PQ}$.

Prepp Team · Answer

$P^2Q$

Prepp Team · Answer

$PQ^2$

Prepp Team · Answer

$P^{1/3} Q^{2/3}$

Understanding Geometric Progression and Mean

Setting up the Geometric Progression

Calculating P: GM of the First \(n\) Terms

Calculating Q: GM of Terms \(a_{n+1}\) to \(a_{3n}\)

Evaluating Option 3: \(\sqrt{PQ}\)

Conclusion

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