# If $(1), (log_yx), (log_zy), (-15log_xz)$ are in Arithmetic Progression...

If $(1), (log_yx), (log_zy), (-15log_xz)$ are in AP then...
(A)$z^3=x$
(B)$x=y^{-1}$
(C)$z^{-3}=y$
(D)$x=y^{-1}=z^3$

I tried many approaches, one of them being:
$1+log_zy=2log_yx$
$log_zz+log_zy=log_yx^2$
$log_z(yz)=log_yx^2$
But the fact that the bases are different is causing the main problem as I don't know how to proceed further. It would be great if anybody could give me a hint for solving such questions.

• Set the second term in the sequence equal to $1+r$. Solve for $y$ in terms of $x$ and $r$. Substitute this expression for $y$ into the 3rd term. Set the 3rd term equal to $1+2r$, and continue on in this manner for the next term. I hope this helps. Nov 26, 2016 at 4:25
• @Αδριανός It got simplified to a cubic equation $6r^3+11r^2+6r+16=0$. I am not able to solve further. Nov 26, 2016 at 4:37
• Might be easier to work backward from the proposed answers. E.g., if $z^3=x$, what does that tell you about $-15\log_xz$? From that, you can work out $\log_yx$ and $\log_zy$, and see if it's consistent. Nov 26, 2016 at 5:24

Let $d$ be the common difference of the $A.P.$
$$log_yx = 1 +d \implies x = y^{1+d}$$ $$log_zy = 1+2d \implies y = z^{1+2d}$$ $$-15log_xz = 1+3d \implies z = x^{\frac{-(1+3d)}{15}}$$ Hence, $$x = y^{1+d} = z^{(1+2d)(1+d)} = x^{\frac{-(1+d)(1+2d)(1+3d)}{15}}$$ $$\implies (1+d)(1+2d)(1+3d) = -15$$ $$\implies 6d^3 +11d^2+6d +16 = 0$$ $$\implies (d+2)(6d^2 - d+8) = 0$$ $$\implies d = -2$$
$$\implies x = y^{-1} = z^{3}$$
• Added some more steps. It's nothing but expanding the terms and then observing that $-2$ is the only real root. I personally solve cubic equations by first looking out for a root by trial and error and then factoring appropriately. Nov 26, 2016 at 6:06