m\limits_{n=1}^{2025} \left( \frac{1}{n} - \frac{1}{n+1} \right) $

This series is telescopic, meaning most terms cancel out with each other:

$ = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{2025} - \frac{1}{2026} \right) $

Simplifying the expression, we find:

$ = \frac{1}{1} - \frac{1}{2026} = 1 - \frac{1}{2026} = \frac{2025}{2026} $

Now, evaluate $\sqrt{2026 \cdot S_{2025}}$:

$ \sqrt{2026 \cdot \frac{2025}{2026}} = \sqrt{2025} = 45 $

For the arithmetic progression with the first term $-p$ and common difference $p$, the sum of the first six terms is given by:

$ \frac{6}{2} [-2p + (6-1)p] = 3(5p) = 15p $

Given that:

$ 15p = 45 $

We find:

$ p = 3 $

To determine the absolute difference between the $20^{\text{th}}$ and $15^{\text{th}}$ terms of the A.P., compute:

$ |A_{20} - A_{15}| = |(-p + 19p) - (-p + 14p)| $

Substituting the value of $p$:

$ = |18p - 13p| = |5p| = 5 \times 5 = 25 $

Thus, the absolute difference is $ \boxed{25} $.

19:T7b1,

$ S_{2025} $ का मान ज्ञात करने के लिए, आंशिक योग की गणना करें

$ S_n = \sum\limits_{n=1}^{2025} \frac{1}{n(n+1)} = \sum\limits_{n=1}^{2025} \left( \frac{1}{n} - \frac{1}{n+1} \right) $

यह श्रेणी दूरबीनी है, जिसका अर्थ है कि अधिकांश पद एक-दूसरे के साथ निरस्त हो जाते हैं:

$ = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{2025} - \frac{1}{2026} \right) $

व्यंजक को सरल करने पर, हम पाते हैं:

$ = \frac{1}{1} - \frac{1}{2026} = 1 - \frac{1}{2026} = \frac{2025}{2026} $

अब, $\sqrt{2026 \cdot S_{2025}}$ का मूल्यांकन क

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