$a^x=b^y=c^z=d^t$ and $a, b, c, d$ are in GP, then $x, y, z, t$ are in

$(a)\;AP\qquad(b)\;GP\qquad(c)\;HP\qquad(d)\;None$

Answer : (c) HP
Explanation : a , b , c , d are in GP
$b=a^r\quad\;c=ar^2\quad\;d=ar^3$
$a^x=b^y=c^z=d^t$
$x\;log a=y\; log b=z\;log c=t\;log d$
$\large\frac{x}{y}=\large\frac{log b}{log a}=\large\frac{log a_{r}}{log a}$
$\large\frac{x}{y}=\large\frac{log a+log r}{log a}$
$\large\frac{x}{y}=1+\large\frac{log\;r}{log\;a}----(1)$
Similarly , $\large\frac{x}{z}=1+2\;\large\frac{log\;r}{log\;a}----(2)$
$\large\frac{x}{t}=1+3\;\large\frac{log\;r}{log\;a}----(3)$
From $\;(1)\;and\;(2)$
$\large\frac{x}{z}=1+2\;(\large\frac{x}{y}-1)$
$\large\frac{1}{z}=\large\frac{1}{x}+2\;(\large\frac{1}{y}-\large\frac{1}{x})$
$\large\frac{1}{z}=\large\frac{2}{y}-\large\frac{1}{x}$
$\large\frac{1}{x}+\large\frac{1}{z}=\large\frac{2}{y}----(4)$
From $(1)\;,(2)\;and\;(3)$
$\large\frac{x}{y}+\large\frac{x}{t}=2+4\;\large\frac{log\;r}{log\;a}$
$=2\;(1+2\;\large\frac{log\;r}{log\;a})=\large\frac{2x}{z}$
$\large\frac{1}{y}+\large\frac{1}{t}=\large\frac{2}{z}----(5)$
$From (4) \;and\;(5)\;$x , y , z , t are in HP.
answered Jan 24, 2014 by