Let S_{n} = frac{1}{2} + frac{1}{6} + frac{1} | vedclass

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(A) $S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} (\frac{1}{k} - \frac{1}{k+1}) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$.$S_{2025} = \frac{202

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) $S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} (\frac{1}{k} - \frac{1}{k+1}) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$.$S_{2025} = \frac{2025}{2026}$.$\sqrt{2026 \cdot S_{2025}} = \sqrt{2026 \cdot \frac{2025}{2026}} = \sqrt{2025} = 45$.For an $A.P.$ with first term $a = -p$ and common difference $d = p$,the sum of the first $6$ terms is $S_{6} = \frac{6}{2} [2(-p) + (6-1)p] = 3[-2p + 5p] = 3(3p) = 9p$.Given $9p = 45$,so $p = 5$.The $n^{	ext{th}}$ term is $A_{n} = a + (n-1)d = -p + (n-1)p = (n-2)p$.The absolute difference between the $20^{	ext{th}}$ and $15^{	ext{th}}$ terms is $|A_{20} - A_{15}| = |(20-2)p - (15-2)p| = |18p - 13p| = |5p|$.Substituting $p = 5$,we get $|5 	imes 5| = 25$.

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