Question

Let \( a_1, a_2, a_3, \ldots \) be in G.P. If \( a_1 \cdot a_2 \cdot a_3 = 64 \) and \( a_1 \cdot a_2 \cdot a_3 \cdot a_4 \cdot a_5 = 32 \), then common ratio is

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{8} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{1}{2} \)
• E. \( \frac{1}{4} \)

Hint:

When products of G.P. terms are given, convert them into powers of \(a\) and \(r\), then divide equations.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem uses a key property of Geometric Progressions: the product of terms symmetrically placed about the center is constant. For an odd number of terms, their product is the middle term raised to the power of the number of terms.
Step 2: Key Formula or Approach:
For a G.P. with `2k+1` terms, the product of the terms is `(middle term)^(2k+1)`.
1. Product of the first 3 terms (a\(_1\), a\(_2\), a\(_3\)): The middle term is a\(_2\). Product = (a\(_2\))\(^3\).
2. Product of the first 5 terms (a\(_1\), ..., a\(_5\)): The middle term is a\(_3\). Product = (a\(_3\))\(^5\).
The common ratio `r` is given by `r = a_{n} / a_{n-1}`.
Step 3: Detailed Explanation:
From the first condition:
\[ a_1 \cdot a_2 \cdot a_3 = 64 \] Using the property for the product of 3 terms, we have:
\[ (a_2)^3 = 64 = 4^3 \] \[ a_2 = 4 \] From the second condition:
\[ a_1 \cdot a_2 \cdot a_3 \cdot a_4 \cdot a_5 = 32 \] Using the property for the product of 5 terms, we have:
\[ (a_3)^5 = 32 = 2^5 \] \[ a_3 = 2 \] Now we have the second term `a_2 = 4` and the third term `a_3 = 2`. The common ratio `r` can be found by dividing a term by its preceding term:
\[ r = \frac{a_3}{a_2} = \frac{2}{4} = \frac{1}{2} \] Step 4: Final Answer:
The common ratio is 1/2.