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Let a1, a2, a3, ..., a49 be in AP such that Σa(4k + 1) for k is 0 to 12 = 416

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Let a1, a2, a3, ..., a49 be in AP such that Σa4k + 1 for k is 0 to 12 = 416 and a9 + a43 = 66. If a12 + a22 + ....+ a172 = 140m, then m is equal to

(a) 66 

(b) 68

(c) 34

(d) 33

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Correct option (c) 34

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