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$x+2$ 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 $S_{\infty} = 6$
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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
by Diamond (44,680 points)

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