Key Concepts and Formulas

1. The $n$-th term of an A.P.: $T_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
2. The sum of the first $n$ terms of an A.P.: $S_n = \frac{n}{2}[2a + (n-1)d]$.

Step-by-Step Solution

Step 1: Formulate equations from the given information. We are given that the 10th term ($T_{10}$) is $\frac{1}{20}$ and the 20th term ($T_{20}$) is $\frac{1}{10}$. Using the formula for the $n$-th term, $T_n = a + (n-1)d$, we can set up two equations: For $n=10$: $T_{10} = a + (10-1)d \implies a + 9d = \frac{1}{20}$ (Equation 1) For $n=20$: $T_{20} = a + (20-1)d \implies a + 19d = \frac{1}{10}$ (Equation 2)

Step 2: Solve for the common difference ($d$). To find $d$, we subtract Equation 1 from Equation 2. This eliminates $a$ and allows us to solve for $d$. $(a + 19d) - (a + 9d) = \frac{1}{10} - \frac{1}{20}$ $10d = \frac{2}{20} - \frac{1}{20}$ $10d = \frac{1}{20}$ Dividing both sides by 10, we get: $d = \frac{1}{20 \times 10} = \frac{1}{200}$

Step 3: Solve for the first term ($a$). Substitute the value of $d = \frac{1}{200}$ into Equation 1: $a + 9\left(\frac{1}{200}\right) = \frac{1}{20}$ $a + \frac{9}{200} = \frac{1}{20}$ To solve for $a$, subtract $\frac{9}{200}$ from both sides. We convert $\frac{1}{20}$ to $\frac{10}{200}$: $a = \frac{10}{200} - \frac{9}{200}$ $a = \frac{1}{200}$ So, the first term is $a = \frac{1}{200}$ and the common difference is $d = \frac{1}{200}$.

Step 4: Calculate the sum of the first 200 terms ($S_{200}$). We use the formula for the sum of the first $n$ terms, $S_n = \frac{n}{2}[2a + (n-1)d]$, with $n=200$, $a=\frac{1}{200}$, and $d=\frac{1}{200}$. $S_{200} = \frac{200}{2}\left[2\left(\frac{1}{200}\right) + (200-1)\left(\frac{1}{200}\right)\right]$ $S_{200} = 100\left[\frac{2}{200} + 199\left(\frac{1}{200}\right)\right]$ $S_{200} = 100\left[\frac{2}{200} + \frac{199}{200}\right]$ $S_{200} = 100\left[\frac{2+199}{200}\right]$ $S_{200} = 100\left[\frac{201}{200}\right]$ $S_{200} = \frac{100 \times 201}{200}$ $S_{200} = \frac{201}{2}$ $S_{200} = 100.5$

Step 5: Match the result with the given options. The calculated sum is $100.5$. This can be written as $100 \frac{1}{2}$. Comparing this with the given options: (A) 100 (B) $100{1 \over 2}$ (C) $50{1 \over 4}$ (D) 50

Our result matches option (B).

Common Mistakes & Tips

• Fraction Arithmetic: Be meticulous with calculations involving fractions. Ensure correct common denominators are used.
• Formula Application: Double-check that the correct formula for the $n$-th term ($a + (n-1)d$) and sum ($S_n$) is applied.
• Systematic Solving: Break down the problem into finding $d$, then $a$, and finally $S_n$. This structured approach minimizes errors.

Summary

The problem required finding the sum of the first 200 terms of an Arithmetic Progression given two terms. By setting up and solving a system of linear equations for the first term ($a$) and common difference ($d$), we found $a = \frac{1}{200}$ and $d = \frac{1}{200}$. Subsequently, using the formula for the sum of an A.P., we calculated $S_{200} = 100.5$, which is equivalent to $100 \frac{1}{2}$.

The final answer is $\boxed{100{1 \over 2}}$ which corresponds to option (B).