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Sum of 1/(sqrt(a_k) + sqrt(a_(k+1))) for an A.P.

This type of question, asked 3 times: 31 Jan 2023, 6 Apr 2023, 5 Apr 2024. All 3 questions below, oldest first.

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  1. Question 1

    31 Jan 2023

    89
    Let a1,a2,…,ana_{1},a_{2},\ldots,a_{n} be in A.P. If a5=2a7a_{5}= 2a_{7} and a11=18a_{11}= 18, then 12(1a10+a11+1a11+a12+⋯+1a17+a18)12\left( \frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+ \cdots +\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}} \right) is equal to

  2. Question 2

    6 Apr 2023

    75
    Let a1,a2,a3…ana_{1},a_{2},a_{3}\ldots a_{n} be nn positive consecutive terms of an arithmetic progression. If d>0d > 0 is its common difference, then
    lim⁡n→∞dn(1a1+a2+1a2+a3+……….+1an−1+an)\lim_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left( \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+ \ldots\ldots\ldots. +\frac{1}{\sqrt{a_{n - 1}}+\sqrt{a_{n}}} \right)

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  3. Question 3

    5 Apr 2024

    76
    If 11+2+12+3+…+199+100=m\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+ \ldots +\frac{1}{\sqrt{99}+\sqrt{100}}= m and 11⋅2+12⋅3+…+199⋅100=n\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+ \ldots +\frac{1}{99 \cdot 100}= n, then the point (m,n)(m,n) lies on the line

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