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Sum of 1/(a_k a_(k+1)) for terms of an A.P., by telescoping

This type of question, asked 8 times: 2021 Paper 9, 2021 Paper 24, 25 Jul 2022, 29 Jul 2022, 30 Jan 2023, 9 Apr 2024, 24 Jan 2025, 28 Jan 2025. All 8 questions below, oldest first.

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  1. Question 1

    2021 Paper 9

    75
    132−1+152−1+172−1+…+1(201)2−1\frac{1}{3^{2}- 1}+\frac{1}{5^{2}- 1}+\frac{1}{7^{2}- 1}+ \ldots +\frac{1}{(201)^{2}- 1} is equal to

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  2. Question 2

    2021 Paper 24

    78
    Let a1,a2,……,a21a_{1},a_{2},\ldots\ldots,a_{21} be an AP such that ∑n=1201anan+1=49\sum_{n= 1}^{20} \frac{1}{a_{n}a_{n+ 1}}=\frac{4}{9}. If the sum of this AP is 189, then a6a16a_{6}a_{16} is equal to:

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  3. Question 3

    25 Jul 2022

    155
    The sum ∑n=1213(4n−1)(4n+3)\sum_{n = 1}^{21} \frac{3}{(4n - 1)(4n + 3)} is equal to

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  4. Question 4

    29 Jul 2022

    64
    If 1(20−a)(40−a)+1(40−a)(60−a)+……+\frac{1}{(20 -a)(40 -a)}+\frac{1}{(40 -a)(60 -a)}+ \ldots\ldots + 1(180−a)(200−a)=1256\frac{1}{(180 -a)(200 -a)}=\frac{1}{256}, then the maximum value of a is :

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  5. Question 5

    30 Jan 2023

    69
    If an=−24n2−16n+15a_{n}=\frac{- 2}{4n^{2}- 16n + 15}, then a1+a2+⋯…+a25a_{1}+a_{2}+ \cdots\ldots +a_{25} is equal to:

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  6. Question 6

    9 Apr 2024

    72
    If the sum of series
    11⋅(1+d)+1(1+d)(1+2 d)+…...+1(1+9 d)(1+10 d)\frac{1}{1 \cdot (1 + d)}+\frac{1}{(1 + d)(1 + 2\text{ }d)}+ \ldots... +\frac{1}{(1 + 9\text{ }d)(1 + 10\text{ }d)}
    is equal to 5, then 50 d50\text{ }d is equal to:

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  7. Question 7

    24 Jan 2025

    54
    Let Sn=12+16+112+120+…S_{n}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+ \ldots upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is 2026 S2025\sqrt{2026{\text{ }S}_{2025}}, then the absolute difference between 20th 20^{\text{th~}} and 15th 15^{\text{th~}} terms of the A.P. is

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  8. Question 8

    28 Jan 2025

    138
    For positive integers nn, if 4an=(n2+5n+6)4a_{n}=\left( n^{2}+ 5n + 6 \right) and Sn=∑k=1n(1ak)S_{n}=\sum_{k = 1}^{n} \left( \frac{1}{a_{k}} \right), then the value of 507 S2025507{\text{ }S}_{2025} is :

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