If the first 3 terms of a GP are a – 1, a + 3 and 3a + 1 can the 4th term be determined?

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The first three terms of a geometric progression are a – 1, a + 3 and 3a + 1. As the square of the 2nd term is equal to the product of the 1st and 3rd terms: (a + 3)^2 = (a - 1)(3a + 1)

=> a^2 + 6a + 9 = 3a^2 - 3a + a - 1

=> 2a^2 - 8a - 10 = 0

=> 2a^2 - 10a + 2a - 10 = 0

=> 2a(a - 5) + 2(a - 5) = 0

=> (2a + 2)(a - 5) = 0

=> a = -1 and a = 5

The terms of the GP are -2, 2, -2 or 4, 8, 16

The 4th term in the first case is 2 and in the second case it is 32.

The 4th term of the geometric series that has a – 1, a + 3 and 3a + 1 as the first three terms can either be 2 or 32.

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