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

To solve the problem, consider a geometric progression (G.P.) defined by terms \( a_1, a_2, a_3, \ldots \). In a G.P., the \( n \)-th term is defined as \( a_n = a_1 \cdot r^{n-1} \) where \( r \) is the common ratio.

Given conditions: 

• The sum of the 6th and 8th terms is 2.
• The product of the 3rd and 5th terms is \(\frac{1}{9}\).

 

Let's translate these into equations using the G.P. term formula:

1. The 6th term, \( a_6 \), is \( a_1 \cdot r^5 \). The 8th term, \( a_8 \), is \( a_1 \cdot r^7 \). We have: \(a_1 \cdot r^5 + a_1 \cdot r^7 = 2\). This simplifies to: \(a_1 \cdot r^5 (1 + r^2) = 2\). [Equation 1]
2. The 3rd term, \( a_3 \), is \( a_1 \cdot r^2 \). The 5th term, \( a_5 \), is \( a_1 \cdot r^4 \). The product is: \((a_1 \cdot r^2) \cdot (a_1 \cdot r^4) = \frac{1}{9}\), which simplifies to: \(a_1^2 \cdot r^6 = \frac{1}{9}\). [Equation 2]

From Equation 2, we can express: \(a_1 \cdot r^3 = \frac{1}{3}\). By squaring both sides, we obtain: \(a_1^2 \cdot r^6 = \frac{1}{9}\), which is already satisfied by Equation 2.

Now, substituting \( a_1 \cdot r^3 = \frac{1}{3} \) into Equation 1:

\(\frac{1}{3} r^2 (1 + r^2) = 2\). Simplify to find \( r^2 \): \(\frac{r^2 (1 + r^2)}{3} = 2\), \(r^2 (1 + r^2) = 6\), \(r^4 + r^2 - 6 = 0\).

This is a quadratic in terms of \( x = r^2 \):

\(x^2 + x - 6 = 0\) Factoring, we find \((x - 2)(x + 3) = 0\), giving solutions \(x = 2\) or \(x = -3\). Since \( x = r^2 \) and must be positive, \(\(r^2 = 2\)\).

Using \( r^2 = 2 \), calculate \( a_2 \) and \( a_4 \):

• \(a_2 = a_1 \cdot r = a_1 \sqrt{2}\),
• \(a_4 = a_1 \cdot r^3 = \frac{1}{3}\sqrt{2}\).

 

Now calculate the desired expression \( 6(a_2 + a_4)(a_4 + a_6) \):

\(a_6 = a_1 \cdot r^5 = \frac{1}{3} \cdot 2\sqrt{2} = \frac{2\sqrt{2}}{3}\)

Substituting:

• \(a_2 + a_4 = a_1 \sqrt{2} + \frac{\sqrt{2}}{3} = \frac{a_1 \cdot 3 \sqrt{2} + \sqrt{2}}{3}\)
• \(a_4 + a_6 = \frac{\sqrt{2}}{3} + \frac{2 \sqrt{2}}{3} = \sqrt{2}\right)\)

Therefore:

\(= 6 \cdot \left(\frac{a_1 \cdot 3 \sqrt{2} + \sqrt{2}}{3}\right) \cdot \sqrt{2}\),

On simplification: \(6 \cdot \frac{3 \sqrt{2}}{3} = 3\).

The answer is 3.