In a increasing geometric series, the sum of the second and the sixth term is `(25)/(2)` and the product of the third and fifth term is 25. then , the sum of ` 4^(th) , 6 ^(th) and 8^(th)` terms to

Options

A. 30
B. 26
C. 35
D. 32

Answer

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

- Second term: \( T_2 = ar \)
- Third term: \( T_3 = ar^2 \)
- Fourth term: \( T_4 = ar^3 \)
- Fifth term: \( T_5 = ar^4 \)
- Sixth term: \( T_6 = ar^5 \)
- Eighth term: \( T_8 = ar^7 \)

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

From the problem, we have two conditions:

1. The sum of the second and sixth terms is \( \frac{25}{2} \):
\[ T_2 + T_6 = ar + ar^5 = ar(1 + r^4) = \frac{25}{2} \]

2. The product of the third and fifth terms is 25:
\[ T_3 \cdot T_5 = (ar^2)(ar^4) = a^2 r^6 = 25 \]

### Step 2: Express \( a \) in terms of \( r \)

From the second equation, we can express \( a^2 \):
\[ a^2 = \frac{25}{r^6} \]
Taking the square root gives:
\[ a = \frac{5}{r^3} \]

### Step 3: Substitute \( a \) into the first equation

Substituting \( a \) into the first equation:
\[ \frac{5}{r^3} r(1 + r^4) = \frac{25}{2} \]
This simplifies to:
\[ \frac{5(1 + r^4)}{r^2} = \frac{25}{2} \]

### Step 4: Clear the fraction

Cross-multiplying gives:
\[ 10(1 + r^4) = 25r^2 \]
This simplifies to:
\[ 10 + 10r^4 = 25r^2 \]
Rearranging gives:
\[ 10r^4 - 25r^2 + 10 = 0 \]

### Step 5: Substitute \( x = r^2 \)

Let \( x = r^2 \), then the equation becomes:
\[ 10x^2 - 25x + 10 = 0 \]

### Step 6: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{25 \pm \sqrt{(-25)^2 - 4 \cdot 10 \cdot 10}}{2 \cdot 10} \]
\[ x = \frac{25 \pm \sqrt{625 - 400}}{20} \]
\[ x = \frac{25 \pm \sqrt{225}}{20} \]
\[ x = \frac{25 \pm 15}{20} \]
Calculating the two possible values:
1. \( x = \frac{40}{20} = 2 \)
2. \( x = \frac{10}{20} = \frac{1}{2} \)

Since \( r^2 \) must be positive and the series is increasing, we take \( r^2 = 2 \) which gives \( r = \sqrt{2} \).

### Step 7: Find \( a \)

Substituting \( r \) back to find \( a \):
\[ a = \frac{5}{r^3} = \frac{5}{(\sqrt{2})^3} = \frac{5}{2\sqrt{2}} = \frac{5\sqrt{2}}{4} \]

### Step 8: Calculate \( T_4 + T_6 + T_8 \)

Now we can find the fourth, sixth, and eighth terms:
\[ T_4 = ar^3 = \frac{5\sqrt{2}}{4} \cdot (\sqrt{2})^3 = \frac{5\sqrt{2}}{4} \cdot 2\sqrt{2} = \frac{10 \cdot 2}{4} = 5 \]
\[ T_6 = ar^5 = \frac{5\sqrt{2}}{4} \cdot (\sqrt{2})^5 = \frac{5\sqrt{2}}{4} \cdot 4\sqrt{2} = \frac{20 \cdot 2}{4} = 10 \]
\[ T_8 = ar^7 = \frac{5\sqrt{2}}{4} \cdot (\sqrt{2})^7 = \frac{5\sqrt{2}}{4} \cdot 8\sqrt{2} = \frac{40 \cdot 2}{4} = 20 \]

Now, summing these terms:
\[ T_4 + T_6 + T_8 = 5 + 10 + 20 = 35 \]

### Final Answer
The sum of the fourth, sixth, and eighth terms is \( 35 \).