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

Concept: In G.P., terms are: \[ a, ar, ar^2, ar^3, ar^4 \]

Step 1: Form first product.
\[ a_1 a_2 a_3 = a \cdot ar \cdot ar^2 = a^3 r^3 = 64 \quad ...(1) \]

Step 2: Form second product.
\[ a_1 a_2 a_3 a_4 a_5 = a^5 r^{10} = 32 \quad ...(2) \]

Step 3: Divide (2) by (1).
\[ \frac{a^5 r^{10}}{a^3 r^3} = \frac{32}{64} \] \[ a^2 r^7 = \frac{1}{2} \quad ...(3) \]

Step 4: Use substitution from (1).
From (1): \[ a^3 r^3 = 64 \Rightarrow (ar)^3 = 64 \Rightarrow ar = 4 \]

Step 5: Solve for \(r\).
\[ a = \frac{4}{r} \] Substitute into (3): \[ \left(\frac{4}{r}\right)^2 r^7 = \frac{1}{2} \] \[ \frac{16}{r^2} \cdot r^7 = \frac{1}{2} \Rightarrow 16r^5 = \frac{1}{2} \] \[ r^5 = \frac{1}{32} \Rightarrow r = \frac{1}{2} \]