Let `I_(1) : (log_(x)2) (log_(2x)2) (log_(2)4x)gt1`
`I_(2) : x^((log_(10)x)^(2)-3(log_(10)x)+1) gt 1000`
and solution of inequality `I_(1)`
is `((1)/(a^(sqrt(a))),(1)/(b))cup(c, a^(sqrt(a)))`
and solution of inequality `I_(2)` is `(d, oo)` then answer the following
Sum of `'d'` term of a `GP` whose common ratio is `(1)/(a)` and first term is `c` is more than

Options

A. `1`
B. `2`
C. `(3)/(2)`
D. `0`

Explanation

To solve the inequalities \( I_1 \) and \( I_2 \) and find the required sum of the GP, we will follow these steps:

### Step 1: Solve Inequality \( I_1 \)

The inequality is given as:
\[ I_1: (\log_x 2)(\log_{2x} 2)(\log_2 4x) > 1 \]

1. **Change of Base**: We can express the logarithms in terms of base 2:
\[ \log_x 2 = \frac{1}{\log_2 x}, \quad \log_{2x} 2 = \frac{1}{\log_2 (2x)} = \frac{1}{\log_2 2 + \log_2 x} = \frac{1}{1 + \log_2 x}, \quad \log_2 4x = \log_2 4 + \log_2 x = 2 + \log_2 x \]

2. **Substituting**: Let \( t = \log_2 x \). Then we have:
\[ I_1: \left(\frac{1}{t}\right)\left(\frac{1}{1+t}\right)(2+t) > 1 \]

3. **Simplifying**: Multiply through by \( t(1+t) \) (assuming \( t > 0 \)):
\[ (2+t) > t(1+t) \]
Simplifying gives:
\[ 2 + t > t + t^2 \implies 2 > t^2 \]
Thus, we find:
\[ t^2 < 2 \implies -\sqrt{2} < t < \sqrt{2} \]

4. **Back Substituting**: Recall \( t = \log_2 x \):
\[ -\sqrt{2} < \log_2 x < \sqrt{2} \]
This translates to:
\[ 2^{-\sqrt{2}} < x < 2^{\sqrt{2}} \]

### Step 2: Solve Inequality \( I_2 \)

The inequality is given as:
\[ I_2: x^{(\log_{10} x)^2 - 3\log_{10} x + 1} > 1000 \]

1. **Taking Logarithm**: Taking logarithm on both sides:
\[ (\log_{10} x)^2 - 3\log_{10} x + 1 > 3 \]
Rearranging gives:
\[ (\log_{10} x)^2 - 3\log_{10} x - 2 > 0 \]

2. **Factoring**: We can factor this quadratic:
\[ (\log_{10} x - 3)(\log_{10} x + 1) > 0 \]

3. **Finding Intervals**: The critical points are \( \log_{10} x = 3 \) and \( \log_{10} x = -1 \):
- \( \log_{10} x < -1 \) or \( \log_{10} x > 3 \)
- This translates to:
\[ x < 10^{-1} \quad \text{or} \quad x > 10^3 \]
Thus:
\[ (0, 0.1) \cup (1000, \infty) \]

### Step 3: Identify Values of \( a, b, c, d \)

From the solutions:
- From \( I_1 \): \( a = 2 \), \( b = 2 \), \( c = 1 \)
- From \( I_2 \): \( d = 1000 \)

### Step 4: Calculate the Sum of the GP

The sum of the first \( d \) terms of a GP with first term \( c \) and common ratio \( \frac{1}{a} \):
\[ S_d = c \frac{1 - r^d}{1 - r} = 1 \cdot \frac{1 - \left(\frac{1}{2}\right)^{1000}}{1 - \frac{1}{2}} = \frac{1 - \left(\frac{1}{2}\right)^{1000}}{\frac{1}{2}} = 2 \left(1 - \frac{1}{2^{1000}}\right) \]

### Final Answer
\[ S_d = 2 \left(1 - \frac{1}{2^{1000}}\right) \]