Question

If $1 , \log _ { y } x , \log _ { z } y , - 15 \log _ { x } z$ are in A.P., then.

• A. $z ^ { 3 } = x$
• B. $x = y ^ { - 1 }$
• C. $z ^ { - 3 } = y$
• D. $x = y ^ { - 1 } = z ^ { 3 }$ (e) All the above

Answer 1

(5)

Sol. Let be the common difference then

$\log _ { y } x = 1 + d$ $\Rightarrow$ $x = y ^ { 1 + d }$

$\log _ { z } y = 1 + 2 d$ $\Rightarrow$ $y = z ^ { 1 + 2 d }$and$- 15 \log _ { x } z = 1 + 3 d$

$\Rightarrow$ $z = x ^ { - ( 1 + 3 d ) / 15 }$

$\therefore$ $x = y ^ { 1 + d } = z ^ { ( 1 + 2 d ) ( 1 + d ) } = x ^ { - ( 1 + d ) ( 1 + 2 d ) ( 1 + 3 d ) / 15 }$

$\Rightarrow$ $( 1 + d ) ( 1 + 2 d ) ( 1 + 3 d ) = - 15$

$\Rightarrow$ $6 d ^ { 3 } + 11 d ^ { 2 } + 6 d + 16 = 0$

$\Rightarrow$ $( d + 2 ) \left( 6 d ^ { 2 } - d + 8 \right) = 0$ $\Rightarrow$ $d = - 2$

[Note that $6 d ^ { 2 } - d + 8 = 0$ has complex roots]

$\therefore$ $x = y ^ { 1 + d } = y ^ { - 1 } , y = z ^ { 1 - 4 } = z ^ { - 3 }$

$\therefore$ $x = \left( z ^ { - 3 } \right) ^ { - 1 } = z ^ { 3 }$ . Also $x = y ^ { - 1 } = z ^ { 3 }$ .