JEE Main | 2018Math miscellaneousHard

Question

Let a1, a2, a3, ..... , a49 be in A.P. such that ∑k=012a4k+1 = 416 and a9 + a43 = 66. If a12+a22+....+a172 = 140m, then m is equal to-

Options

A.

68

B.

34

C.

33

D.

66

Solution

 ∑k=012a4k+1 =416
⇒132 [2a1 + 48d] = 416
⇒a1 + 24d  = 3 2 ..... (1)
a9 + a43 = 66⇒ 2a1 + 50d = 66 ..... (2)
                          2a1 + 48d = 64 by ...... (1)
                     ________________________
                    d = 1 and a1 = 8
⇒ 140m =∑r =117ar2∑r=117[8+(r-1).1]2
⇒ 140m =∑r =117(r + 7)2
⇒ 140m =∑r =124r2-∑r=17r2
⇒ 140m = 24.25.496-7.8.156
⇒ 140m = 7.8.56[105 - 3]
⇒ 140m = 280.17 ⇒ m = 34

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