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Let `a_1, a_2, a_3...a_49` be in AP such that `sum_(k=0)^12(a_4k+1)=416` and `a_9+a_43=66` If `a_1^2+a_2^2+...+a_17^2=140m` then m is equal to (1) 66 (2) 68 (3) 34 (4) 33

A

66

B

68

C

34

D

33

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The correct Answer is:
C
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