Question

Let $a, \frac{3}{4}, ar^{2}, ar^{3}, \dots$ be in G.P. where $r>0$. If the product of the first four terms is $\frac{3^{6}}{4^{5}}$, then $a$ is equal to ________.

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$
• E. 1

Hint:

$n$-th term of G.P. is $ar^{n-1}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
We are given a geometric progression (G.P.) and information about the product of its first four terms and its common ratio. We need to find the first term, 'a'.
Step 2: Key Formula or Approach
The first four terms of the G.P. are \(a, ar, ar^2, ar^3\).
The product (P) of these four terms is:
\[ P = (a) \times (ar) \times (ar^2) \times (ar^3) = a^4 r^{0+1+2+3} = a^4 r^6 \] We are given \(P = \frac{4}{9}\) and \(r = \frac{2}{3}\). We will substitute these values into the formula and solve for \(a\).
Step 3: Detailed Explanation
1. Set up the equation using the given values.
\[ a^4 r^6 = P \] \[ a^4 \left(\frac{2}{3}\right)^6 = \frac{4}{9} \] 2. Isolate the term \(a^4\).
\[ a^4 = \frac{4/9}{(2/3)^6} = \frac{4}{9} \times \frac{1}{(2/3)^6} = \frac{4}{9} \times \left(\frac{3}{2}\right)^6 \] 3. Simplify the expression.
Express the numbers as powers of their prime factors (2 and 3) to simplify the calculation.
\[ a^4 = \frac{2^2}{3^2} \times \frac{3^6}{2^6} \] Using the laws of exponents (\(\frac{x^m}{x^n} = x^{m-n}\)):
\[ a^4 = 2^{2-6} \times 3^{6-2} = 2^{-4} \times 3^4 = \frac{3^4}{2^4} \] \[ a^4 = \left(\frac{3}{2}\right)^4 \] 4. Solve for a.
Taking the fourth root of both sides:
\[ a = \pm \frac{3}{2} \] The problem states that \(r>0\), but typically in such problems 'a' is also assumed to be positive unless otherwise specified. Given the options, we choose the positive root. \[ a = \frac{3}{2} \] Step 4: Final Answer
The value of a is \(\frac{3}{2}\).