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If $a_1=1/\left ( 2^{*}5 \right ),a_2=1/\left ( 5^{*}8 \right ),a_3=1/\left ( 8^{*}11 \right ),\dots\dots,$ then $a_1+a_2+\dots\dots+a_{100}$ is

  1. $25/151$
  2. $1/2$
  3. $1/4$
  4. $111/55$
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Given that,

  • $a_1=\frac{1}{2 \ast 5}$  
  • $a_2=\frac{1}{5 \ast 8}$
  • $a_3=\frac{1}{8 \ast 11}$
  • $a_4=\frac{1}{11 \ast 14}$
  • $\vdots \quad \vdots \quad \vdots \quad \vdots$

   We can generalize the term:

  • $a_n=\dfrac{1}{(3n-1)(3n-2)};\forall n \geq 1$

Now,   $a_1+a_2+a_3+a_4+ \ldots +a_{100} = \frac{1}{2 \ast 5}+\frac{1}{5 \ast 8}+\frac{1}{8 \ast 11}+\frac{1}{11 \ast 14}+ \ldots +\frac{1}{299 \ast 302}$

$\qquad \qquad = \dfrac{1}{3} \left[\dfrac{3}{2 \ast 5}+\dfrac{3}{5 \ast 8}+\dfrac{3}{8 \ast 11}+\dfrac{3}{11 \ast 14}+ \ldots +\dfrac{3}{299 \ast 302}\right]$

$\qquad \qquad = \dfrac{1}{3}\left[\dfrac{5-2}{2 \ast 5}+\dfrac{8-5}{5 \ast 8}+\dfrac{11-8}{8 \ast 11}+\dfrac{14-11}{11 \ast 14}+ \ldots +\dfrac{302-299}{299 \ast 302}\right]$

$\qquad \qquad = \dfrac{1}{3}\left[\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+ \ldots +\dfrac{1}{299}-\dfrac{1}{302}\right]$

$\qquad \qquad  = \dfrac{1}{3}\left[\dfrac{1}{2}-\dfrac{1}{302}\right] = \dfrac{1}{3}\left[\dfrac{151-1}{302}\right] = \dfrac{1}{3}\times\dfrac{150}{302}=\dfrac{25}{151}$

$\therefore$ The value of $a_1+a_2+a_3+a_4+ \ldots +a_{100}$ is  $\dfrac{25}{151}.$

Correct Answer $:\text{A}$

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