# Progressions

• Apr 30th 2010, 08:05 PM
yshridhar
Progressions
If 1, log x (base y), log y (base z), -15 log z (base x)
are in AP, then
a) z^3 = x
b) x = y^-1
c) z^-3 = y
d) all of these

Regards
Sreedhar
• Apr 30th 2010, 11:53 PM
hollywood
Let the common difference be d. Then the progression is 1, d+1, 2d+1, and 3d+1, so
$\displaystyle x=y^{d+1}$
$\displaystyle y=z^{2d+1}$
$\displaystyle z=x^{-(3d+1)/15}\text{ or }x=z^{-15/(3d+1)}$
Combining the first two gives $\displaystyle x=z^{(2d+1)(d+1)}$, and equating exponents gives (3d+1)(2d+1)(d+1)=-15, with the only solution d=-2.

- Hollywood
• May 1st 2010, 06:48 AM
Soroban
Hello, yshridhar!

Lovely work, Hollywood!

Quote:

$\displaystyle \text{If }\;1,\;\log_y(x),\;\log_z(y),\;-15\log_x(z)\,\text{ are in AP, then:}$

. . $\displaystyle (a)\;z^3 = x \qquad (b)\;x = y^{-1} \qquad (c)\;z^{-3} = y \qquad (d)\;\text{all of these}$

Hollywood had: .$\displaystyle \begin{Bmatrix}[1] & x &=& y^{d+1} \\ [2] & y &=& z^{2d+1} \\ [3] &z &=& x^{-\frac{3x+1}{15}}\end{Bmatrix}\quad\text{ and }\:d =- 2$

Then we have:

. . $\displaystyle [1]\;\;x \:=\:y^{-1} \;\;{\color{blue}(b)}$

. . $\displaystyle [2]\;\;y \:=\:z^{-3} \quad\Rightarrow\quad z^{-3} \:=\:y \;\;{\color{blue}(c)}$

. . $\displaystyle [3]\;z \:=\:x^{\frac{1}{3}} \quad\Rightarrow\quad z^3\:=\:x\;\;{\color{blue}(a)}$

Answer: .$\displaystyle {\color{blue}(d)\;\text{ all of these}}$

• May 1st 2010, 07:31 PM
yshridhar
Thanks alot Hollywood for such a simple solution. I am banging with logarithms formulae for this sum.
Regards
Sridhar