The numbers a1,a2,a3,... form an arithmetic sequence with a1≠a2. The three numbers a1,a2,a6 form a geometric sequence in that orde

Question

The numbers a1,a2,a3,... form an arithmetic sequence with a1≠a2. The three numbers a1,a2,a6 form a geometric sequence in that order. Determine all possible positive integers k for which the three numbers a1,a4,ak also form a geometric sequence in that order.

Amity · Answer

All the positive integers k for which three numbers a1,a4,ak form a  geometric sequence  in order is  37 .    It is given that,   a₁ , a₂ , a₃   form an  an  arithmetic sequence , a₁ ≠  a₂, a₁ , a₂ , a₆  form geometric sequence.   For finding, all possible  positive  integers k for which the three numbers a₁, a₄, ak  form a geometric sequence  Then,   a₁ , a₂ , a₃  , .................. aₙ    is  AP  a₁ ≠  a₂   (common difference is not zero)  aₙ = a + (n - 1) d  a₁ , a₂ , a₆  form GP  a₂²   = a₁ a₆ (a₁ + d)²    = a₁ (a₁ + 5d)  a₁²  + d² + 2a₁d    = a₁² + a₁5d  d² + 2a₁d  =  a₁5d  d  + 2a₁  = 5a₁ (as d is not equal to 0)   3a₁  = d  d = 3a₁  a₄  = a₁  + 3(3a₁)  = 10a₁  ak  = a₁ + (k - 1)(3a₁)    = a + 3ka₁  - 3a₁    = a₁ (3k - 2)  Forming the  geometric  sequence, a₁, a₄, ak      a₄²  =  a₁ . (ak)  (10a₁)²  = a₁ a₁(3k - 2)  3k - 2 = 100  3k = 102    k = 34     To learn more about  geometric sequence : https://brainly.com/question/1509142  #SPJ4

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