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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