Let ${a_1}, {a_2}, \dots, {a_{49}}$ be in $A.P.$ such that $\sum_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $\sum_{r = 1}^{17} a_r^2 = 140m$,then $m = \dots$

  • A
    $68$
  • B
    $34$
  • C
    $33$
  • D
    $66$

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