If the sum of the series $\frac{1}{1 \cdot(1+d)} + \frac{1}{(1+d)(1+2d)} + \dots + \frac{1}{(1+9d)(1+10d)}$ is equal to $5$,then $50d$ is equal to:

  • A
    $20$
  • B
    $5$
  • C
    $15$
  • D
    $10$

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