Question

Let a₁, a₂, a₃, …. be a G.P. of increasing positive numbers. Let the sum of its 6^th and 8^th terms be 2 and the product of its 3rd and 5th terms be \(\frac{1}{9}\) .Then 6 (a₂ + a₄) (a₄ + a₆) is equal to

• A. 2
• B. \(2\sqrt2\)
• C. \(3\sqrt3\)
• D. 3

Answer 1

The Correct Option is D

\( a_3 \cdot a_5 = \frac{1}{9} \)

\( \Rightarrow ar^2 \cdot ar^4 = \frac{1}{9} \)

\( \Rightarrow (ar^3)^2 = \frac{1}{9} \)

\( \Rightarrow ar^3 = \frac{1}{3} \)    ...(i)

\( a_6 + a_8 = 2 \)

\( \Rightarrow ar^5 + ar^7 = 2 \)

\( \Rightarrow ar^3 (r^2 + r^4) = 2 \)

\( \Rightarrow \frac{1}{3} r^2 (1 + r^2) = 2 \)

\( \Rightarrow r^2 (1 + r^2) = 2 \times 3 \)

\( \Rightarrow r^2 = 2 \Rightarrow r = \sqrt{2} \)

\( a = \frac{1}{3} \times \frac{1}{r^3} \)

\( = \frac{1}{3} \times \frac{1}{2\sqrt{2}} = \frac{1}{6\sqrt{2}} \)

\( 6(a_2 + a_4)(a_4 + a_6) \)

\( \Rightarrow 6(ar + ar^3)(ar^3 + ar^5) \)

\( \Rightarrow 6 \left( \frac{ar^3}{r^2} + \frac{1}{3} \right) \left( \frac{1}{3} + \frac{1}{3} r^2 \right) = 3 \)