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If a₁, a₂, , a_n is a sequence of positive numbers which are in A.P. with common difference d and a₁+a₄+a₇+ .+a₁₆=147 then a₁+a₁₆=M and a₁+a₆+a₁₁+a₁₆=N Maximum value of a₁ a₂ . a₁₆= ( S W )¹⁶ (where S and W are coprime), then:

Options

  1. AM=49
  2. BN=98
  3. CS=49
  4. DW=2

Correct answer

D. W=2

Step-by-step solution

Using, A.M. G.M. a_i 16 (a₁ a₂ a₁₆ )^ 1 / 16 a₁ a₂ a₁₆ ( a₁+a₂+ a₁₆ 16 )¹⁶ ( 392 16 )¹⁶ ( 49 2 )¹⁶ S=49 and W=2 6 2 (2 a+5 3 d)=1472 a+15 d=49 ...(1) 2 a+15 d=M=49 Also, 4 2 (2 a+3 5 d)=N ; (2 a+15 d) 2=N N=98

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