Question

A geometric series has firs term of 4 and common ratio r, where 0 < r < 1. Given that the first, second and forth terms of this geometric series form three consecutive terms of an arithmetic series, show that r3 − 2r + 1 = 0 and find the value of r.

Answer 1

To show that r^3 - 2r + 1 = 0, we established that the first, second, and fourth terms of the geometric series are also consecutive terms of an arithmetic series. After equating terms and simplifying the expression, we derived the cubic equation and found that r equals 1.

Step-by-step explanation:

A geometric series has a first term of 4 and a common ratio r, where 0 < r < 1. If the first, second, and fourth terms form an arithmetic series, the sequence is as follows:

• First term (a1): 4
• Second term (a2): 4r
• Fourth term (a4): 4r3

The common difference d of the arithmetic series is the difference between the second and first term:

d = a2 - a1 = 4r - 4

Since a4 is the third consecutive term in the arithmetic sequence, we can express it as:

a4 = a2 + d = 4r + (4r - 4)

Simplify and set equal to the fourth term of the geometric sequence:

4r3 = 4r + 4r - 4

Divide through by 4:

r3 - 2r + 1 = 0

This is the cubic equation to solve to find the common ratio r.

To find r, we can factor the equation:

(r - 1)2(r + 1) = 0

Since 0 < r < 1, the value for r is 1.