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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial