Key Concepts and Formulas

• Telescoping Series: A series of the form $\sum_{k=1}^{n} (f(k) - f(k+1))$ which simplifies to $f(1) - f(n+1)$.
• Arithmetic Progression (A.P.):
  • The $n^{th}$ term is $A_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
  • The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.
• Partial Fraction Decomposition: Used to express a rational function as a sum of simpler fractions. For $\frac{1}{x(x+1)}$, it is $\frac{1}{x} - \frac{1}{x+1}$.

Step-by-Step Solution

Step 1: Evaluate the sum $S_{2025}$ We are given $S_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \ldots$ up to $n$ terms. We can observe that the general term of the series can be written as $\frac{1}{k(k+1)}$. So, $S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)}$. Using partial fraction decomposition, we write $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$. Substituting this back into the sum, we get: $S_n = \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right)$ This is a telescoping series. Let's expand the first few terms and the last term: $S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$ The intermediate terms cancel out, leaving: $S_n = 1 - \frac{1}{n+1}$ Now, we need to find $S_{2025}$ by substituting $n = 2025$: $S_{2025} = 1 - \frac{1}{2025+1} = 1 - \frac{1}{2026} = \frac{2026 - 1}{2026} = \frac{2025}{2026}$

Step 2: Calculate the value of $\sqrt{2026 \cdot S_{2025}}$ We are given an expression involving $S_{2025}$ that relates to the sum of the A.P. $\sqrt{2026 \cdot S_{2025}} = \sqrt{2026 \cdot \frac{2025}{2026}}$ The $2026$ in the numerator and denominator cancel out: $\sqrt{2026 \cdot S_{2025}} = \sqrt{2025}$ We know that $\sqrt{2025} = 45$.

Step 3: Determine the value of $p$ using the A.P. sum Let the A.P. have the first term $a = -p$ and the common difference $d = p$. The sum of the first six terms of this A.P. is given by $S_6 = \frac{6}{2}[2a + (6-1)d]$. Substituting $a = -p$ and $d = p$: $S_6 = 3[2(-p) + 5(p)] = 3[-2p + 5p] = 3(3p) = 9p$ We are given that this sum is equal to $\sqrt{2026 \cdot S_{2025}}$, which we found to be 45. So, we have the equation: $9p = 45$ Dividing both sides by 9 to solve for $p$: $p = \frac{45}{9} = 5$

Step 4: Find the absolute difference between the 20th and 15th terms of the A.P. The $n^{th}$ term of the A.P. is $A_n = a + (n-1)d$. With $a = -p$ and $d = p$, the formula becomes $A_n = -p + (n-1)p$. We need to find $|A_{20} - A_{15}|$. First, let's find the 20th term, $A_{20}$: $A_{20} = -p + (20-1)p = -p + 19p = 18p$ Next, let's find the 15th term, $A_{15}$: $A_{15} = -p + (15-1)p = -p + 14p = 13p$ Now, we compute the absolute difference: $|A_{20} - A_{15}| = |18p - 13p| = |5p|$ Substitute the value of $p = 5$ that we found: $|5p| = |5 \cdot 5| = |25| = 25$

Common Mistakes and Tips

• Ensure careful application of partial fraction decomposition to correctly identify the terms in the telescoping series.
• Double-check the formulas for the $n^{th}$ term and the sum of an A.P. to avoid errors.
• When calculating the difference between terms, remember that the difference between the $m^{th}$ and $n^{th}$ term of an A.P. is $|(m-n)d|$. In this case, it is $|(20-15)p| = |5p|$.

Summary

The problem required us to first evaluate the sum $S_{2025}$ by recognizing it as a telescoping series after applying partial fraction decomposition. We then used this value to determine the sum of the first six terms of the given arithmetic progression. By equating this sum to the calculated value and using the properties of the A.P. (first term $-p$ and common difference $p$), we found the value of $p$. Finally, we calculated the absolute difference between the 20th and 15th terms of the A.P. using the formula for the $n^{th}$ term.

The final answer is $\boxed{25}$.