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Let a 1 , a 2 , a 3 , … … , a 49 be in A . P . such that Σ k ⁡ = 0 1 2 a 4 k + 1 = 416 and a 9 + a 43 = 66 . If a 1 2 + a 2 2 + … + a 17 2 = 140 m , then m is equal to:

Options

  1. A33
  2. B66
  3. C68
  4. D34

Correct answer

D. 34

Step-by-step solution

a 1 + a 5 +   .   .   .   + a 49 = 416 Let common difference =   d 13 a 1   + 4 d 1 + 2 +   .   .   .   + 12 = 416 13 a 1   + 4 d 12 13 2 = 416 13 a 1   + 24 d 13 = 416 a 1 + 24 d = 32 .   .   .   ( i ) a 9 + a 43 = 66 a 1 + 8 d + a 1 + 42 d = 66 2 a 1 + 50 d = 66 a 1 + 25 d = 33 .   .   .   ( i i ) a 1 = 8       ;     d = 1 ∑ i = 1 17 a i 2 = 140 m ⇒ a 2 + a + d 2 +   . .   . &#1

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