Question

Given that \( a, \frac{3}{4} a r^2, a r^3 \) are in G.P. Product of first four terms = \( \frac{3^6}{4^3} \), then find \( a \):

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

Hint:

In geometric progressions, use the common ratio to solve for unknowns. Remember that the product of terms follows a specific formula.

Answer 1

The Correct Option is B

The terms \( a, \frac{3}{4} a r^2, a r^3 \) form a geometric progression (G.P.). In a G.P., the ratio of consecutive terms is constant. Therefore, we can write: \[ \frac{\frac{3}{4} a r^2}{a} = \frac{a r^3}{\frac{3}{4} a r^2} \] Simplifying both sides: \[ \frac{3}{4} r^2 = \frac{4}{r} \] This leads to: \[ r^3 = \frac{4}{3} \] Taking the cube root of both sides: \[ r = \frac{4^{1/3}}{3^{1/3}} \] We are also given that the product of the first four terms is: \[ a \cdot \frac{3}{4} a r^2 \cdot a r^3 \cdot a r^4 = \frac{3^6}{4^3} \] Solving this equation will lead us to find \( a = \frac{3}{4} \).