Question

Given: x-1;x+1;x...as the first 3 terms of a geometric sequence

Answer 1

Answer:

see explanation

Step-by-step explanation:

(1)

The common ratio r of a geometric sequence is

r = $\frac{a_{2} }{x_{1} }$ = $\frac{a_{3} }{a_{2} }$ , so

$\frac{x+1}{x-1}$ = $\frac{x}{x+1}$ ( cross- multiply )

(x + 1)² = x(x - 1) ← expand both sides

x² + 2x + 1 = x² - x ( subtract x² - x from both sides )

3x + 1 = 0 ( subtract 1 from both sides )

3x = - 1 ( divide both sides by 3 )

x = - $\frac{1}{3}$

(2)

r = $\frac{x}{x+1}$ = $\frac{-\frac{1}{3} }{-\frac{1}{3}+1 }$ = $\frac{-\frac{1}{3} }{\frac{2}{3} }$ = - $\frac{1}{2}$

(3)

If - 1 < r < 1 then the sequence will converge

r = - $\frac{1}{2}$ meets this criteria , thus the sequence is convergent