If `sin theta,sqrt(2) (sintheta + 1),6 sin theta +6` are in GP, than the fifth term is

Options

A. 81
B. `81sqrt(2)`
C. 162
D. `162sqrt(2)`

Explanation

To solve the problem where \( \sin \theta, \sqrt{2}(\sin \theta + 1), 6 \sin \theta + 6 \) are in geometric progression (GP), we will follow these steps:

### Step 1: Set up the condition for GP
For three numbers \( a, b, c \) to be in GP, the condition is:
\[ b^2 = ac \]
In our case:
\[ (\sqrt{2}(\sin \theta + 1))^2 = \sin \theta (6 \sin \theta + 6) \]

### Step 2: Expand both sides
Expanding the left-hand side:
\[ (\sqrt{2}(\sin \theta + 1))^2 = 2(\sin \theta + 1)^2 = 2(\sin^2 \theta + 2\sin \theta + 1) = 2\sin^2 \theta + 4\sin \theta + 2 \]

Expanding the right-hand side:
\[ \sin \theta (6 \sin \theta + 6) = 6 \sin^2 \theta + 6 \sin \theta \]

### Step 3: Set the expanded forms equal to each other
Now we set the two expansions equal to each other:
\[ 2\sin^2 \theta + 4\sin \theta + 2 = 6\sin^2 \theta + 6\sin \theta \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 2\sin^2 \theta + 4\sin \theta + 2 - 6\sin^2 \theta - 6\sin \theta = 0 \]
This simplifies to:
\[ -4\sin^2 \theta - 2\sin \theta + 2 = 0 \]
Multiplying through by -1:
\[ 4\sin^2 \theta + 2\sin \theta - 2 = 0 \]

### Step 5: Solve the quadratic equation
We can simplify this equation:
\[ 2\sin^2 \theta + \sin \theta - 1 = 0 \]
Using the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = 1, c = -1 \):
\[ \sin \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
This gives us:
\[ \sin \theta = \frac{2}{4} = \frac{1}{2} \quad \text{(valid)} \quad \text{or} \quad \sin \theta = \frac{-4}{4} = -1 \quad \text{(not valid)} \]

### Step 6: Find the common ratio \( R \)
Now, we find the common ratio \( R \):
\[ R = \frac{\sqrt{2}(\sin \theta + 1)}{\sin \theta} = \frac{\sqrt{2}(\frac{1}{2} + 1)}{\frac{1}{2}} = \frac{\sqrt{2} \cdot \frac{3}{2}}{\frac{1}{2}} = 3\sqrt{2} \]

### Step 7: Find the fifth term
The first term \( A = \sin \theta = \frac{1}{2} \) and the common ratio \( R = 3\sqrt{2} \). The fifth term \( T_5 \) is given by:
\[ T_5 = A \cdot R^4 = \frac{1}{2} \cdot (3\sqrt{2})^4 \]
Calculating \( (3\sqrt{2})^4 \):
\[ (3\sqrt{2})^4 = 3^4 \cdot (\sqrt{2})^4 = 81 \cdot 4 = 324 \]
Thus:
\[ T_5 = \frac{1}{2} \cdot 324 = 162 \]

### Final Answer
The fifth term is \( \boxed{162} \).

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