Let `a_1,a_2,a_3,.......` 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 `3^(rd)` and `5^(th)` terms be (1/2) . then `6(a_2 +a_4)(a_4+a_6)` is equal to

Options

A. 2sqrt2`
B. `3sqrt3`
C. `2`
D. `3`

Answer

To solve the problem, we will denote the first term of the geometric progression (G.P.) as \( a \) and the common ratio as \( r \). The terms of the G.P. can be expressed as follows:

- \( a_1 = a \)
- \( a_2 = ar \)
- \( a_3 = ar^2 \)
- \( a_4 = ar^3 \)
- \( a_5 = ar^4 \)
- \( a_6 = ar^5 \)
- \( a_7 = ar^6 \)
- \( a_8 = ar^7 \)

### Step 1: Set up the equations based on the problem statement

From the problem, we have two conditions:
1. The sum of the 6th and 8th terms is 2:
\[ a_6 + a_8 = ar^5 + ar^7 = 2 \]
This can be factored as:
\[ ar^5(1 + r^2) = 2 \quad \text{(Equation 1)} \]

2. The product of the 3rd and 5th terms is \( \frac{1}{2} \):
\[ a_3 \cdot a_5 = ar^2 \cdot ar^4 = a^2 r^6 = \frac{1}{2} \quad \text{(Equation 2)} \]

### Step 2: Solve Equation 2 for \( a^2 \)

From Equation 2:
\[ a^2 r^6 = \frac{1}{2} \]
We can express \( a^2 \) in terms of \( r \):
\[ a^2 = \frac{1}{2r^6} \quad \text{(Equation 3)} \]

### Step 3: Substitute \( a^2 \) into Equation 1

Now we can substitute \( a \) from Equation 3 into Equation 1. First, we need to express \( a \) in terms of \( r \):
\[ a = \sqrt{\frac{1}{2r^6}} = \frac{1}{\sqrt{2} r^3} \]

Substituting \( a \) into Equation 1:
\[ \frac{1}{\sqrt{2} r^3} r^5 (1 + r^2) = 2 \]
This simplifies to:
\[ \frac{r^2 (1 + r^2)}{\sqrt{2}} = 2 \]
Multiplying both sides by \( \sqrt{2} \):
\[ r^2 (1 + r^2) = 2\sqrt{2} \]

### Step 4: Rearranging the equation

Rearranging gives:
\[ r^4 + r^2 - 2\sqrt{2} = 0 \]
Let \( x = r^2 \), then we have:
\[ x^2 + x - 2\sqrt{2} = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-1 \pm \sqrt{1 + 8\sqrt{2}}}{2} \]

### Step 6: Find \( a_2, a_4, a_6 \)

Now we need to find \( a_2, a_4, a_6 \):
- \( a_2 = ar = \frac{1}{\sqrt{2} r^3} r = \frac{1}{\sqrt{2} r^2} \)
- \( a_4 = ar^3 = \frac{1}{\sqrt{2} r^3} r^3 = \frac{1}{\sqrt{2}} \)
- \( a_6 = ar^5 = \frac{1}{\sqrt{2} r^3} r^5 = \frac{r^2}{\sqrt{2}} \)

### Step 7: Calculate \( 6(a_2 + a_4)(a_4 + a_6) \)

Now we can calculate:
\[ 6(a_2 + a_4)(a_4 + a_6) \]
Substituting the values:
\[ = 6\left(\frac{1}{\sqrt{2} r^2} + \frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}} + \frac{r^2}{\sqrt{2}}\right) \]
This simplifies to:
\[ = 6\left(\frac{1 + r^2}{\sqrt{2} r^2}\right)\left(\frac{1 + r^2}{\sqrt{2}}\right) \]
\[ = \frac{6(1 + r^2)^2}{2r^2} = \frac{3(1 + r^2)^2}{r^2} \]

### Final Step: Substitute \( r^2 \) back to find the value

Using the value of \( r^2 \) from the quadratic solution, we can find the final value.