Question

Let $G_1, G_2, G_3$ be geometric means between $l$ and $n$, where $l$ and $n$ are positive real numbers. Then the common ratio is

• A. $\frac{n}{l}$
• B. $(\frac{n}{l})^{1/2}$
• C. $(\frac{n}{l})^{1/3}$
• D. $(\frac{n}{l})^{1/4}$
• E. $\frac{n^2}{l^2}$

Hint:

The common ratio for $k$ geometric means between $a$ and $b$ is $r = (\frac{b}{a})^{\frac{1}{k+1}}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When \( k \) geometric means are inserted between two numbers \( a \) and \( b \), they form a Geometric Progression (G.P.) with a total of \( k+2 \) terms. The first term is \( a \) and the last term is \( b \). In this problem, we have 3 geometric means between \( l \) and \( n \).
Step 2: Key Formula or Approach:
The sequence is \( l, G_1, G_2, G_3, n \).
This is a G.P. with:
First term, \( a_1 = l \).
Total number of terms = 3 (means) + 2 (endpoints) = 5 terms.
The last term, \( a_5 = n \).
The formula for the n-th term of a G.P. is \( a_k = a_1 \cdot r^{(k-1)} \). We will use this to find the common ratio \( r \).
Step 3: Detailed Explanation:
Using the formula for the 5th term (\( a_5 \)):
\[ a_5 = a_1 \cdot r^{(5-1)} \] Substitute the given values \( a_1 = l \) and \( a_5 = n \):
\[ n = l \cdot r^4 \] Our goal is to find the common ratio, \( r \). We need to solve this equation for \( r \).
Divide both sides by \( l \):
\[ \frac{n}{l} = r^4 \] To find \( r \), take the fourth root of both sides:
\[ r = \sqrt[4]{\frac{n}{l}} \] This can also be written using fractional exponents:
\[ r = \left(\frac{n}{l}\right)^{\frac{1}{4}} \] Step 4: Final Answer:
The common ratio of the G.P. is \( \left(\frac{n}{l}\right)^{\frac{1}{4}} \).