Let `t_(n)` be the nth term of an A.P. If `sum_(r = 1)^(10^(99)) a_(2r) = 10^(100) and sum_(r = 1)^(10^(99)) a_(2r - 1) = 10^(99)`, then the common difference of A.P. is

Options

A. 1
B. 10
C. 9
D. `10^(99)`

Answer

To find the common difference of the arithmetic progression (A.P.) given the conditions in the problem, we can follow these steps:

### Step 1: Define the terms of the A.P.
Let the first term of the A.P. be \( a \) and the common difference be \( d \). The \( n \)-th term of the A.P. can be expressed as:
\[ t_n = a + (n-1)d \]

### Step 2: Express the sums of even and odd indexed terms
We need to find the sums of the even indexed terms and the odd indexed terms up to \( n = 10^{99} \).

1. **Sum of even indexed terms**:
The even indexed terms are \( t_2, t_4, t_6, \ldots, t_{2 \cdot 10^{99}} \).
The sum of these terms can be expressed as:
\[ S_{\text{even}} = t_2 + t_4 + t_6 + \ldots + t_{2 \cdot 10^{99}} = (a + d) + (a + 3d) + (a + 5d) + \ldots + (a + (2 \cdot 10^{99} - 1)d) \]
This is an arithmetic series with \( 10^{99} \) terms, first term \( t_2 = a + d \) and last term \( t_{2 \cdot 10^{99}} = a + (2 \cdot 10^{99} - 1)d \).
The sum can be calculated as:
\[ S_{\text{even}} = \frac{n}{2} \times (\text{first term} + \text{last term}) = \frac{10^{99}}{2} \times \left( (a + d) + \left( a + (2 \cdot 10^{99} - 1)d \right) \right) \]
Simplifying this gives:
\[ S_{\text{even}} = 10^{98} \times (2a + (10^{99})d) = 10^{100} \]

2. **Sum of odd indexed terms**:
The odd indexed terms are \( t_1, t_3, t_5, \ldots, t_{2 \cdot 10^{99} - 1} \).
The sum of these terms can be expressed as:
\[ S_{\text{odd}} = t_1 + t_3 + t_5 + \ldots + t_{2 \cdot 10^{99} - 1} = a + (a + 2d) + (a + 4d) + \ldots + (a + (2 \cdot 10^{99} - 2)d) \]
This is also an arithmetic series with \( 10^{99} \) terms, first term \( t_1 = a \) and last term \( t_{2 \cdot 10^{99} - 1} = a + (2 \cdot 10^{99} - 2)d \).
The sum can be calculated as:
\[ S_{\text{odd}} = \frac{10^{99}}{2} \times \left( a + \left( a + (2 \cdot 10^{99} - 2)d \right) \right) \]
Simplifying this gives:
\[ S_{\text{odd}} = 10^{98} \times (2a + (10^{99} - 1)d) = 10^{99} \]

### Step 3: Set up the equations
From the sums we have:
1. \( 10^{98} \times (2a + 10^{99}d) = 10^{100} \)
2. \( 10^{98} \times (2a + (10^{99} - 1)d) = 10^{99} \)

### Step 4: Simplify the equations
Dividing both equations by \( 10^{98} \):
1. \( 2a + 10^{99}d = 10^{2} \) (Equation 1)
2. \( 2a + (10^{99} - 1)d = 10^{1} \) (Equation 2)

### Step 5: Subtract the equations
Subtract Equation 2 from Equation 1:
\[ (2a + 10^{99}d) - (2a + (10^{99} - 1)d) = 10^{2} - 10^{1} \]
This simplifies to:
\[ d = 10^{2} - 10^{1} = 100 - 10 = 90 \]

### Step 6: Conclusion
The common difference \( d \) of the A.P. is:
\[ \boxed{90} \]