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Let-a-b-c-be-in-AP-and-a-lt-1-b-lt-1-c-lt-1-If-x-1-a-a-2-to-y-1-b-b-2-to-z-1-c-c-2-to-then-x-y-z-are-in-

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Question Number 56419 by gunawan last updated on 16/Mar/19
Let a, b, c be in AP and ∣a∣<1, ∣b∣<1, ∣c∣<1. If x = 1+a+a^2 +... to ∞ y = 1+b+b^2 +... to ∞ z = 1+c+c^2 +... to ∞ then x, y, z are in
$$\mathrm{Let}\:{a},\:{b},\:{c}\:\mathrm{be}\:\mathrm{in}\:\mathrm{AP}\:\mathrm{and}\:\mid{a}\mid<\mathrm{1},\:\mid{b}\mid<\mathrm{1},\:\mid{c}\mid<\mathrm{1}.\:\mathrm{If} \\ $$$${x}\:\:=\:\:\mathrm{1}+{a}+{a}^{\mathrm{2}} +…\:\mathrm{to}\:\infty\: \\ $$$${y}\:\:=\:\:\mathrm{1}+{b}+{b}^{\mathrm{2}} +…\:\mathrm{to}\:\infty\: \\ $$$${z}\:\:=\:\:\mathrm{1}+{c}+{c}^{\mathrm{2}} +…\:\mathrm{to}\:\infty\:\: \\ $$$$\mathrm{then}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{in} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19
x=(1/(1−a)) y=(1/(1−b)) z=(1/(1−c)) 1−a=(1/x)→a=1−(1/x) b=1−(1/y) c=1−(1/z) 2b=a+c 2(1−(1/y))=1−(1/x)+1−(1/z) (2/y)=(1/x)+(1/z) so x,y and z in H.P
$${x}=\frac{\mathrm{1}}{\mathrm{1}−{a}}\:\:{y}=\frac{\mathrm{1}}{\mathrm{1}−{b}}\:\:\:{z}=\frac{\mathrm{1}}{\mathrm{1}−{c}} \\ $$$$\mathrm{1}−{a}=\frac{\mathrm{1}}{{x}}\rightarrow{a}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$${b}=\mathrm{1}−\frac{\mathrm{1}}{{y}} \\ $$$${c}=\mathrm{1}−\frac{\mathrm{1}}{{z}} \\ $$$$\mathrm{2}{b}={a}+{c} \\ $$$$\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{{y}}\right)=\mathrm{1}−\frac{\mathrm{1}}{{x}}+\mathrm{1}−\frac{\mathrm{1}}{{z}} \\ $$$$\frac{\mathrm{2}}{{y}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}\:\:{so}\:{x},{y}\:{and}\:{z}\:{in}\:{H}.{P} \\ $$

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