Given that the sum of infinity of geometric series a−2ar+4ar^2 −8ar^3 +...a(−2r)^(n−1) +...is 3 and the sum of infinity of geometric series a+ar+ar^2 +ar^3 +...ar^(n−1) +... is k, find the range of values of k.


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Question Number 115946 by ZiYangLee last updated on 29/Sep/20

$Given that the sum of infinity of geometric$ $series a - 2 a r + 4 {a r}^{2} - 8 {a r}^{3} + . . . a {(- 2 r)}^{n - 1} + . . . is 3$ $and the sum of infinity of geometric series$ $a + a r + {a r}^{2} + {a r}^{3} + . . . {a r}^{n - 1} + . . . is k,$ $find the range of values of k .$
	Answered by Dwaipayan Shikari last updated on 29/Sep/20
	$a−2ar+4ar^2 −8ar^3 +....=a((1/(1+2r))) (a/(1+2r))=3⇒3+6r=a⇒r=((a−3)/6) a(1+r+r^2 +...)=(a/(1−r))=(a/(1−((a−3)/6)))=(a/(9−a))=k k{0,(1/8),(2/7),(1/2),(4/5),(5/4),2,(7/2),8} 0≤k≤8$
	$a - 2 ar + {4 ar}^{2} - {8 ar}^{3} + . . . . = a (\frac{1}{1 + 2 r})$ $\frac{a}{1 + 2 r} = 3 \Rightarrow 3 + 6 r = a \Rightarrow r = \frac{a - 3}{6}$ $a (1 + r + r^{2} + . . .) = \frac{a}{1 - r} = \frac{a}{1 - \frac{a - 3}{6}} = \frac{a}{9 - a} = k$ $k {0, \frac{1}{8}, \frac{2}{7}, \frac{1}{2}, \frac{4}{5}, \frac{5}{4}, 2, \frac{7}{2}, 8}$ $0 ⩽ k ⩽ 8$
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