$OABCD$ is a pentagon in which the sides $OA$ and $CB$ are parallel and the sides $OD$ and $AB$ are parallel. Also,it is given that $\frac{OA}{CB}=2$,$\frac{OD}{AB}=\frac{1}{3}$. If $\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$,then $\vec{AD}+\vec{OC}+\vec{DC}=$

  • A
    $\vec{d}-\vec{a}$
  • B
    $\frac{1}{2}\vec{a}+3\vec{d}$
  • C
    $\frac{1}{2}\vec{a}+2\vec{d}$
  • D
    $6\vec{d}$

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