In a quadrilateral $PQRS$, $A$ divides $SR$ in the ratio $1:3$ and $B$ is the mid-point of $PR$. If $3SR - QR - 3PS - PQ = kAB$, then $k=$

  • A
    $2$
  • B
    $4$
  • C
    $6$
  • D
    $8$

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Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the midpoints of the diagonals $AC$ and $BD$ respectively and $(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})= k \overrightarrow{FE}$,then $k$ is equal to

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