Question

$x+2$ and $x^{2} - 4$ are the first and second terms of a geometric series respectively.

Calculate:

1. the value(s) of x for which the series converges
2. x if $S_{\infty} = 6$

Answer 1

1. r= (x^2-4)/(x+2) = (x-2)(x+2)/(x+2)

                          r = x-2

for convergence,   -1<r<1

                            = -1<x-2<1

                            = 1< x < 3

2. sum to infinity= a/(1-r)

                     6= (x+2)/(1-(x-2))

                     6= (x+2)/(3-x)

                   6(3-x)= x+2

                        18-6x=x+2

                          7x= 16

                                x= 16/7