Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant.
  • General Term: $A_k = A_1 + (k-1)d$, where $A_1$ is the first term and $d$ is the common difference.
  • Sum of first $n$ terms: $S_n = \frac{n}{2}(2A_1 + (n-1)d)$ or $S_n = \frac{n}{2}(A_1 + A_n)$.
• Inequalities with Reciprocals: For positive numbers $a$ and $b$, if $a < b$, then $\frac{1}{a} > \frac{1}{b}$. Conversely, if $a > b$, then $\frac{1}{a} < \frac{1}{b}$.

Step-by-Step Solution

Step 1: Define the A.P. and extract initial information. We are given that $\frac{1}{x_1}, \frac{1}{x_2}, ..., \frac{1}{x_n}$ are in A.P. Let $A_k = \frac{1}{x_k}$. We are given $x_1 = 4$, so the first term of the A.P. is $A_1 = \frac{1}{x_1} = \frac{1}{4}$. We are given $x_{21} = 20$, so the 21st term of the A.P. is $A_{21} = \frac{1}{x_{21}} = \frac{1}{20}$. We need to find $n$, the least positive integer such that $x_n > 50$. This implies $A_n = \frac{1}{x_n} < \frac{1}{50}$. Finally, we need to calculate the sum $S_n = \sum_{i=1}^n A_i$.

Step 2: Calculate the common difference ($d$) of the A.P. We use the formula for the general term of an A.P.: $A_k = A_1 + (k-1)d$. For $k=21$, we have $A_{21} = A_1 + (21-1)d$. Substituting the known values: $\frac{1}{20} = \frac{1}{4} + 20d$ To solve for $d$, rearrange the equation: $20d = \frac{1}{20} - \frac{1}{4}$ Find a common denominator (20): $20d = \frac{1}{20} - \frac{5}{20}$ $20d = \frac{1 - 5}{20}$ $20d = -\frac{4}{20}$ $20d = -\frac{1}{5}$ Divide by 20: $d = -\frac{1}{5 \times 20} = -\frac{1}{100}$ The common difference is $d = -\frac{1}{100}$.

Step 3: Determine the least positive integer $n$ for which $x_n > 50$. The condition $x_n > 50$ is equivalent to $\frac{1}{x_n} < \frac{1}{50}$ because $x_n$ must be positive (since $x_n>50$). So, $A_n < \frac{1}{50}$. Using the general term formula $A_n = A_1 + (n-1)d$: $A_n = \frac{1}{4} + (n-1)\left(-\frac{1}{100}\right)$ $A_n = \frac{1}{4} - \frac{n-1}{100}$ Substitute this into the inequality $A_n < \frac{1}{50}$: $\frac{1}{4} - \frac{n-1}{100} < \frac{1}{50}$ To eliminate the denominators, multiply the entire inequality by 100: $100 \left(\frac{1}{4}\right) - 100 \left(\frac{n-1}{100}\right) < 100 \left(\frac{1}{50}\right)$ $25 - (n-1) < 2$ $25 - n + 1 < 2$ $26 - n < 2$ Subtract 26 from both sides: $-n < 2 - 26$ $-n < -24$ Multiply by -1 and reverse the inequality sign: $n > 24$ Since $n$ is the least positive integer satisfying $n > 24$, we have $n = 25$.

Step 4: Calculate the sum of the first $n$ terms of the A.P. We need to find $S_n = \sum_{i=1}^n A_i$ with $n=25$, $A_1 = \frac{1}{4}$, and $d = -\frac{1}{100}$. Using the sum formula $S_n = \frac{n}{2}(2A_1 + (n-1)d)$: $S_{25} = \frac{25}{2}\left(2 \times \frac{1}{4} + (25-1)\left(-\frac{1}{100}\right)\right)$ $S_{25} = \frac{25}{2}\left(\frac{1}{2} + 24\left(-\frac{1}{100}\right)\right)$ $S_{25} = \frac{25}{2}\left(\frac{1}{2} - \frac{24}{100}\right)$ Simplify $\frac{24}{100}$ to $\frac{6}{25}$: $S_{25} = \frac{25}{2}\left(\frac{1}{2} - \frac{6}{25}\right)$ Find a common denominator for the fractions inside the parenthesis (50): $S_{25} = \frac{25}{2}\left(\frac{25}{50} - \frac{12}{50}\right)$ $S_{25} = \frac{25}{2}\left(\frac{25 - 12}{50}\right)$ $S_{25} = \frac{25}{2}\left(\frac{13}{50}\right)$ Multiply the fractions: $S_{25} = \frac{25 \times 13}{2 \times 50}$ $S_{25} = \frac{25 \times 13}{2 \times 2 \times 25}$ Cancel out the 25: $S_{25} = \frac{13}{2 \times 2}$ $S_{25} = \frac{13}{4}$

Common Mistakes & Tips

• Inequality Reversal: When dealing with reciprocals of inequalities, remember to reverse the inequality sign if the terms are positive. This is crucial when converting $x_n > 50$ to $A_n < \frac{1}{50}$.
• Fraction Arithmetic: Be extremely careful with calculations involving fractions. Ensure a common denominator is used for addition/subtraction and simplify fractions whenever possible to make calculations easier.
• "Least Positive Integer": Pay close attention to this wording. It means after solving an inequality like $n > 24$, you must select the smallest integer that satisfies it, which is 25.

Summary The problem involves an arithmetic progression formed by the reciprocals of a sequence $x_i$. We first identified the first term and the 21st term of this A.P. Using these, we calculated the common difference. Then, we translated the condition $x_n > 50$ into an inequality involving the terms of the A.P. and solved for $n$, finding the least integer value. Finally, we used the formula for the sum of an A.P. to compute the sum of the first $n$ terms.

The final answer is $\boxed{\frac{13}{4}}$.