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If $1,\;{\log _y}x,\;{\log _z}y,\; - 15{\log _x}z$ are in $A.P.$, then
${z^3} = x$
$x = {y^{ - 1}}$
${z^{ - 3}} = y$
All the above
Solution
(d) Let $d$ be the common difference
then ${\log _y}x = 1 + d$
$ \Rightarrow $$x = {y^{1 + d}}$
${\log _z}y = 1 + 2d$
$ \Rightarrow $$y = {z^{1 + 2d}}$
and $ – 15{\log _x}z = 1 + 3d$
$ \Rightarrow $$z = {x^{ – (1 + 3d)/15}}$
$\therefore $$x = {y^{1 + d}} = {z^{(1 + 2d)(1 + d)}} = {x^{ – (1 + d)(1 + 2d)(1 + 3d)/15}}$
$ \Rightarrow $$(1 + d)(1 + 2d)(1 + 3d) = – 15$
$ \Rightarrow $$6{d^3} + 11{d^2} + 6d + 16 = 0$
$ \Rightarrow $$(d + 2)(6{d^2} – d + 8) = 0$
$ \Rightarrow $$d = – 2$
[Note that $6{d^2} – d + 8 = 0$ has complex roots]
$\therefore $ $x = {y^{1 + d}} = {y^{ – 1}},\;y = {z^{1 – 4}} = {z^{ – 3}}$
$\therefore $$x = {({z^{ – 3}})^{ – 1}} = {z^3}$.
Also $x = {y^{ – 1}} = {z^3}$.
