Finding the common difference and hence, the sum of an A.P Find the sum to $25$ terms of an A.P with the first four terms as $1, \log_yx, \log_zy,-15\log_x z$.


My attempt:
I started out with,

$2\log_yx = 1+\log_zy$

and,

$2\log_zy = \log_yx -15\log_xz$

Further simplifying the equations led me nowhere. The work was getting too tedious. And since the exam in which I was supposed to solve this question only gives exactly 180 sec to solve this problem, I thought there might be some other, smarter way.

Any help would be appreciated.
Hint: we have $$\frac{\ln(x)}{\ln(y)}=1+d$$ and $$\frac{\ln(y)}{\ln(z)}=1+2d$$ and $$\frac{\ln(z)}{\ln(x)}=\frac{1+2d}{-15}$$ then we get
$$\frac{(1+2d)(1+3d)}{-15}=\frac{\ln(y)}{\ln(x)}=\frac{1}{1+d}$$ so we obtain
$$(1+2d)(1+3d)(1+d)=-15$$
Can you solve this equation?
Hint: we have $$\frac{\ln(x)}{\ln(y)}=1+d$$ and $$\frac{\ln(y)}{\ln(z)}=1+2d$$ and $$\frac{\ln(z)}{\ln(x)}=\frac{1+2d}{-15}$$ then we get
$$\frac{(1+2d)(1+3d)}{-15}=\frac{\ln(y)}{\ln(x)}=\frac{1}{1+d}$$ so we obtain
$$(1+2d)(1+3d)(1+d)=-15$$
Can you solve this equation?