Question

If the product of five consecutive terms of a G.P. is $\frac{243}{32}$, then the middle term is:

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

Hint:

Always look for patterns in the numbers. Seeing 243 and 32 should immediately trigger the thought of \( 3^5 \) and \( 2^5 \). If the product involves 5 terms, you are almost certainly looking for a fifth power.

Answer 1

The Correct Option is B

Concept: For problems involving the product of terms in a G.P., choosing the terms symmetrically around the center simplifies the algebra significantly because the common ratio terms cancel out during multiplication.

Step 1: Assume the five consecutive terms in a symmetric form.
Let the terms be: \[ \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2 \] In this setup, \( a \) is explicitly the middle term of the sequence.

Step 2: Set up the product equation based on the problem statement.
The product is given as \( \frac{243}{32} \): \[ \left( \frac{a}{r^2} \right) \cdot \left( \frac{a}{r} \right) \cdot (a) \cdot (ar) \cdot (ar^2) = \frac{243}{32} \] Multiplying the terms: \[ \frac{a \cdot a \cdot a \cdot a \cdot a}{r^2 \cdot r \cdot \frac{1}{r} \cdot \frac{1}{r^2}} = \frac{243}{32} \] \[ a^5 = \frac{243}{32} \]

Step 3: Solve for the middle term \( a \).
Identify the base for the power of 5: \[ 243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5 \] \[ 32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5 \] So, we have: \[ a^5 = \left( \frac{3}{2} \right)^5 \] Taking the fifth root of both sides: \[ a = \frac{3}{2} \]