The numbers a1, a2, a3, . . . form an arithmetic sequence with a1 not equal 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.
Answers
Given : a₁ , a₂ , a₃ ____ form an an arithmetic sequence
a₁ ≠ a₂
a₁ , a₂ , a₆ form geometric sequence
To Find : all possible positive integers k for which the three numbers a₁, a₄, ak form a geometric sequence
Solution:
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
as d ≠ 0
=> d + 2a₁ = 5a₁
=> 3a₁ = d
=> d = 3a₁
a₄ = a₁ + 3(3a₁) = 10a₁
ak = a₁ + (k - 1)(3a₁) = a + 3ka₁ - 3a₁ = a₁ (3k - 2)
a₁, a₄, ak form a geometric sequence
=> a₄² = a₁ . (ak)
=> (10a₁)² = a₁ a₁(3k - 2)
=> 3k - 2 = 100
=> 3k = 102
=> k = 34
Value of k is 34
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