If `(1 + 3 + 5 + .... " upto n terms ")/(4 + 7 + 10 + ... " upto n terms") = (20)/(7 " log"_(10)x) and n = log_(10)x + log_(10) x^((1)/(2)) + log_(10) x^((1)/(4)) + log_(10) x^((1)/(8)) + ... + oo`, then x is equal to

Options

A. `10^(3)`
B. `10^(5)`
C. `10^(6)`
D. `10^(7)`

Answer

To solve the given problem step by step, we will break it down into manageable parts.

### Step 1: Understanding the Given Information
We have the equation:
\[ \frac{1 + 3 + 5 + \ldots \text{ (up to n terms)}}{4 + 7 + 10 + \ldots \text{ (up to n terms)}} = \frac{20}{7 \log_{10} x} \]
and we need to find \( x \) given:
\[ n = \log_{10} x + \log_{10} x^{1/2} + \log_{10} x^{1/4} + \log_{10} x^{1/8} + \ldots \]

### Step 2: Simplifying the Expression for \( n \)
The expression for \( n \) can be simplified using the properties of logarithms:
\[ n = \log_{10} x + \log_{10} x^{1/2} + \log_{10} x^{1/4} + \log_{10} x^{1/8} + \ldots \]
This is a geometric series where the first term \( a = \log_{10} x \) and the common ratio \( r = \frac{1}{2} \).

The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting our values:
\[ n = \frac{\log_{10} x}{1 - \frac{1}{2}} = 2 \log_{10} x \]

### Step 3: Finding the Sums of the Series
Next, we need to find the sums of the series in the numerator and the denominator.

**Numerator:**
The series \( 1 + 3 + 5 + \ldots \) is an arithmetic series where:
- First term \( a = 1 \)
- Common difference \( d = 2 \)

The sum of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} (2a + (n - 1)d) \]
Thus,
\[ S_n = \frac{n}{2} (2 \cdot 1 + (n - 1) \cdot 2) = \frac{n}{2} (2 + 2n - 2) = n^2 \]

**Denominator:**
The series \( 4 + 7 + 10 + \ldots \) is also an arithmetic series where:
- First term \( a = 4 \)
- Common difference \( d = 3 \)

Using the same formula:
\[ S_n = \frac{n}{2} (2 \cdot 4 + (n - 1) \cdot 3) = \frac{n}{2} (8 + 3n - 3) = \frac{n}{2} (3n + 5) \]

### Step 4: Setting Up the Equation
Now substituting these sums into our original equation:
\[ \frac{n^2}{\frac{n}{2} (3n + 5)} = \frac{20}{7 \log_{10} x} \]
Simplifying the left side:
\[ \frac{2n}{3n + 5} = \frac{20}{7 \log_{10} x} \]

### Step 5: Substituting \( n = 2 \log_{10} x \)
We substitute \( n \) with \( 2 \log_{10} x \):
\[ \frac{2(2 \log_{10} x)}{3(2 \log_{10} x) + 5} = \frac{20}{7 \log_{10} x} \]
Cross-multiplying gives:
\[ 14 \log_{10} x \cdot 4 \log_{10} x = 20(3(2 \log_{10} x) + 5) \]
This simplifies to:
\[ 56 (\log_{10} x)^2 = 20(6 \log_{10} x + 5) \]
\[ 56 (\log_{10} x)^2 = 120 \log_{10} x + 100 \]
Rearranging gives:
\[ 56 (\log_{10} x)^2 - 120 \log_{10} x - 100 = 0 \]

### Step 6: Solving the Quadratic Equation
Using the quadratic formula:
\[ \log_{10} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 56, b = -120, c = -100 \):
\[ \log_{10} x = \frac{120 \pm \sqrt{(-120)^2 - 4 \cdot 56 \cdot (-100)}}{2 \cdot 56} \]
Calculating the discriminant:
\[ \sqrt{14400 + 22400} = \sqrt{36800} = 192 \]
Thus,
\[ \log_{10} x = \frac{120 \pm 192}{112} \]
Calculating the two possible values:
1. \( \log_{10} x = \frac{312}{112} = \frac{78}{28} = 2.7857 \)
2. \( \log_{10} x = \frac{-72}{112} = -0.6429 \) (not valid since log cannot be negative)

### Step 7: Finding \( x \)
Using \( \log_{10} x = \frac{78}{28} \):
\[ x = 10^{\frac{78}{28}} = 10^{2.7857} \approx 10^{5} \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{10^5} \]