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STD 11 - 8. sequence and series — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 8. sequence and series1 Mark
MCQ
 If $\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+1012}\right) $ $ -\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots+\frac{1}{2024 \cdot 2023}\right) $

$ =\frac{1}{2024}, $ then $\alpha$ is equal to-

  • A
    $1367$
  • B
    $1058$
  • C
    $1056$
  • ✓
    $1011$
✓

Answer

Correct option: D.
$1011$
d
$ \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) $ - $ \left\{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right)\right\}=\frac{1}{2024} $

$\Rightarrow $ $ \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) $ $ -\left\{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\ldots+\frac{1}{2023}\right. $ $ \left.-\frac{1}{2024}-2\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2022}\right)\right\}=\frac{1}{2024} $

$\Rightarrow $ $ \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) $ $ -\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{2023}\right) $ $ \frac{1}{2024}+\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{1011}\right)=\frac{1}{2024} $

$\Rightarrow $ $ \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012} $ = $ \frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023} $

$\Rightarrow $ $ \alpha=1011$

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