Given {a^x} = {b^y} = {c^z} = {d^u} and a, b, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let ${a^x} = {b^y} = {c^z} = {d^u} = k$ (say).Then $a = {k^{1/x}}, b = {k^{1/y}}, c = {k^{1/z}}, d = {k^{1/u}}$.Since $a, b, c, d$ are in $G.P.$,t

Vedclass · Accepted Answer

$H.P.$

Vedclass · Answer

(C) Let ${a^x} = {b^y} = {c^z} = {d^u} = k$ (say).Then $a = {k^{1/x}}, b = {k^{1/y}}, c = {k^{1/z}}, d = {k^{1/u}}$.Since $a, b, c, d$ are in $G.P.$,their logarithms are in $A.P.$ or we can use the property of $G.P.$: ${b^2} = ac$ and ${c^2} = bd$.Substituting the values,we get ${k^{2/y}} = {k^{1/x}} \cdot {k^{1/z}} = {k^{(1/x + 1/z)}}$.This implies $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$,which means $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$Therefore,$x, y, z$ are in $H.P.$Similarly,for $b, c, d$ in $G.P.$,we get $\frac{2}{z} = \frac{1}{y} + \frac{1}{u}$,implying $y, z, u$ are in $H.P.$Thus,$x, y, z, u$ are in $H.P.$

Given ${a^x} = {b^y} = {c^z} = {d^u}$ and $a, b, c, d$ are in $G.P.$,then $x, y, z, u$ are in

Similar Questions

If the $5^{th}$ term of a Harmonic Progression $(H.P.)$ is $1/45$ and the $11^{th}$ term is $1/69$,then its $16^{th}$ term is...

If $\log (x + z) + \log (x + z - 2y) = 2\log (x - z),$ then $x, y, z$ are in

If $\ln(a+c), \ln(c-a), \ln(a-2b+c)$ are in $A.P.$,then

Let $f(x) = x + 1/2$. Then,the number of real values of $x$ for which the three unequal terms $f(x), f(2x), f(4x)$ are in $HP$ is

If $a, b, c$ are in $H.P.$,then the value of $\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module