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
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
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The common ratio for $k$ geometric means between $a$ and $b$ is $r = (\frac{b}{a})^{\frac{1}{k+1}}$.
Updated On: Apr 28, 2026
- $\frac{n}{l}$
- $(\frac{n}{l})^{1/2}$
- $(\frac{n}{l})^{1/3}$
- $(\frac{n}{l})^{1/4}$
- $\frac{n^2}{l^2}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
If $k$ geometric means are inserted between $l$ and $n$, the total number of terms in the G.P. is $k + 2$.
Step 2: Analysis
Here, $k = 3$ ($G_1, G_2, G_3$), so the total terms are 5. The 1st term $a = l$ and the 5th term $a_5 = n$.
Step 3: Calculation
$a_5 = ar^4 \implies n = l \cdot r^4$. $r^4 = \frac{n}{l} \implies r = (\frac{n}{l})^{1/4}$. Final Answer: (D)
If $k$ geometric means are inserted between $l$ and $n$, the total number of terms in the G.P. is $k + 2$.
Step 2: Analysis
Here, $k = 3$ ($G_1, G_2, G_3$), so the total terms are 5. The 1st term $a = l$ and the 5th term $a_5 = n$.
Step 3: Calculation
$a_5 = ar^4 \implies n = l \cdot r^4$. $r^4 = \frac{n}{l} \implies r = (\frac{n}{l})^{1/4}$. Final Answer: (D)
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