a,b,c are in G.P. If a^x =b^y =c^z prove that (1/x),(1/z) are in A.P.


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Question Number 144271 by 7770 last updated on 24/Jun/21

$a, b, c a r e i n G . P . I f a^{x} = b^{y} = c^{z}$ $p r o v e t h a t \frac{1}{x}, \frac{1}{z} a r e i n A . P .$
Commented by Rasheed.Sindhi last updated on 24/Jun/21

$C o r r e c t p l e a s e :$ $p r o v e t h a t \frac{1}{x}, \frac{1}{y}, \frac{1}{z} a r e i n A . P .$
	Answered by Rasheed.Sindhi last updated on 24/Jun/21

	$L e t a^{x} = b^{y} = c^{z} = k$ $x = \frac{\log k}{\log a} \Rightarrow \frac{1}{x} = \frac{\log a}{\log k}$ $y = \frac{\log k}{\log b} \Rightarrow \frac{1}{y} = \frac{\log b}{\log k}$ $x = \frac{\log k}{\log c} \Rightarrow \frac{1}{z} = \frac{\log c}{\log k}$ $\frac{1}{y} - \frac{1}{x} \overset{?}{=} \frac{1}{z} - \frac{1}{y}$ $\frac{\log b}{\log k} - \frac{\log a}{\log k} \overset{?}{=} \frac{\log c}{\log k} - \frac{\log b}{\log k}$ $∵ a, b, c a r e i n G P$ $∴ L e t b = a r & c = {a r}^{2}$ $\frac{\log (a r)}{\log k} - \frac{\log a}{\log k} \overset{?}{=} \frac{\log ({a r}^{2})}{\log k} - \frac{\log (a r)}{\log k}$ $\frac{\log a + \log r - \log a}{\log k} \overset{?}{=} \frac{\log a + 2 \log r - \log a - \log r}{\log k}$ $\frac{\log r}{\log k} = \frac{\log r}{\log k}$ $∴ \frac{1}{x}, \frac{1}{y}, \frac{1}{z} a r e i n A P$
	Answered by Rasheed.Sindhi last updated on 24/Jun/21

	$L e t b = a r & c = {a r}^{2}$ $a^{x} = b^{y} = c^{z} \Rightarrow a^{x} = {(a r)}^{y} = {({a r}^{2})}^{z}$ $x \log a = y (\log a + \log r) = z (\log a + 2 \log r) = k (s a y)$ $x = \frac{k}{\log a} \Rightarrow \frac{1}{x} = \frac{\log a}{k}$ $y = \frac{k}{\log a + \log r} \Rightarrow \frac{1}{y} = \frac{\log a + \log r)}{k}$ $z = \frac{k}{\log a + 2 \log r} \Rightarrow \frac{1}{z} = \frac{\log a + 2 \log r}{k}$ $\frac{1}{y} - \frac{1}{x} \overset{?}{=} \frac{1}{z} - \frac{1}{y}$ $\frac{\log a + \log r)}{k} - \frac{\log a}{k} \overset{?}{=} \frac{\log a + 2 \log r}{k} - \frac{\log a + \log r)}{k}$ $\frac{\log r}{k} = \frac{\log r}{k}$ $∴ \frac{1}{x}, \frac{1}{y}, \frac{1}{z} a r e i n A P$
	Answered by Rasheed.Sindhi last updated on 24/Jun/21

	$a^{x} = b^{y} = c^{z}$ $a = b^{y / x} = c^{z / x}$ $b = a r & c = {a r}^{2}$ $a = a^{y / x} r^{y / x} = a^{z / x} r^{2 z / x}$ $r^{y / x} = a^{1 - \frac{y}{x}} = a^{\frac{x - y}{x}}$ $r = a^{\frac{x - y}{y}} . . . . . . . . . . . . . . (i)$ $a = a^{z / x} r^{2 z / x}$ $r^{2 z / x} = a^{1 - \frac{z}{x}} = a^{\frac{x - z}{x}}$ $r = a^{\frac{x - z}{2 z}} . . . . . . . . . . . . . . . . . . (i i)$ $F r o m (i) & (i i)$ $r = a^{\frac{x - y}{y}} = a^{\frac{x - z}{2 z}} \Rightarrow \frac{x - y}{y} = \frac{x - z}{2 z}$ $2 x z - 2 y z = x y - y z$ $2 x z = x y + y z$ $D i v i d i n g b y x y z :$ $\frac{2}{y} = \frac{1}{z} + \frac{1}{x}$ $\frac{1}{y} = \frac{\frac{1}{x} + \frac{1}{z}}{2}$ $∴ \frac{1}{x}, \frac{1}{y}, \frac{1}{z} a r e i n A P$
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