Let $S_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \ldots$ up to $n$ terms. If the sum of the first six terms of an A.P. with first term $-p$ and common difference $p$ is $\sqrt{2026 S_{2025}}$, then the absolute difference between $20^{\text{th}}$ and $15^{\text{th}}$ terms of the A.P. is

Step 1: Find the general term of the series $S_n$

First, we need to find the general term of the given series $S_n$. We can observe that the denominators are products of consecutive integers: $1 \times 2$, $2 \times 3$, $3 \times 4$, $4 \times 5$, and so on. So, the $k^{\text{th}}$ term, $T_k$, can be written as $\frac{1}{k(k+1)}$. This term can be decomposed using partial fractions into $\frac{1}{k} - \frac{1}{k+1}$.

Step 2: Calculate the sum $S_n$

Now, we calculate the sum $S_n$ by summing the general terms. This is a telescoping series, where most terms cancel out. The sum simplifies to $1 - \frac{1}{n+1}$, which can be written as $\frac{n}{n+1}$.

Step 3: Calculate $S_{2025}$

Using the formula for $S_n$ we just derived, we can find $S_{2025}$ by substituting $n=2025$. This gives us $S_{2025} = \frac{2025}{2026}$.

Step 4: Find the sum of the first six terms of the A.P.

For the A.P., the first term is $a = -p$ and the common difference is $d = p$. The sum of the first six terms, $S_6$, is given by the formula $S_n = \frac{n}{2}(2a + (n-1)d)$. Substituting the values, we get $S_6 = 9p$. We are given that $S_6 = \sqrt{2026 S_{2025}}$. Substituting the value of $S_{2025}$, we find that $9p = 45$, which means $p=5$.

Step 5: Calculate the absolute difference between the 20th and 15th terms

We need to find the absolute difference between the $20^{\text{th}}$ and $15^{\text{th}}$ terms of the A.P. The $n^{\text{th}}$ term of an A.P. is $a_n = a + (n-1)d$. So, $a_{20} = a + 19d$ and $a_{15} = a + 14d$. The difference is $a_{20} - a_{15} = 5d$. Since $d=p$ and we found $p=5$, the difference is $5 \times 5 = 25$. The absolute difference is $|25| = 25$.

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