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.

Answer 1

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: brainly.com/question/1509142

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