Let a, b, c, p, q and r be positive real numb | vedclass

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

Question

(C) Let $a^{p} = b^{q} = c^{r} = k$. Since $a, b, c$ are positive real numbers,we can write $a = k^{1/p}$,$b = k^{1/q}$,and $c = k^{1/r}$. Given that

Vedclass · Accepted Answer

$p, q, r$ are in $H.P.$

Vedclass · Answer

(C) Let $a^{p} = b^{q} = c^{r} = k$. Since $a, b, c$ are positive real numbers,we can write $a = k^{1/p}$,$b = k^{1/q}$,and $c = k^{1/r}$. Given that $a, b, c$ are in $GP$,we have $\frac{b}{a} = \frac{c}{b}$. Substituting the values of $a, b, c$,we get $\frac{k^{1/q}}{k^{1/p}} = \frac{k^{1/r}}{k^{1/q}}$. This implies $k^{(1/q - 1/p)} = k^{(1/r - 1/q)}$. Equating the exponents,we get $\frac{1}{q} - \frac{1}{p} = \frac{1}{r} - \frac{1}{q}$,which simplifies to $\frac{2}{q} = \frac{1}{p} + \frac{1}{r}$. This shows that $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in $A.P.$ Therefore,$p, q, r$ are in $H.P.$

Let $a, b, c, p, q$ and $r$ be positive real numbers such that $a, b$ and $c$ are in $GP$ and $a^{p} = b^{q} = c^{r}$. Then,

Similar Questions

If $\frac{x + y}{2}, y, \frac{y + z}{2}$ are in $H.P.$,then $x, y, z$ are in

If $x_1, x_3$ are the roots of $A x^2 - 4 x + 1 = 0$ and $x_2, x_4$ are the roots of $B x^2 - 6 x + 1 = 0$ such that $x_1, x_2, x_3, x_4$ are in harmonic progression,then $\frac{B+A}{B-A} = $

If $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n}$ is the harmonic mean between $a$ and $b$,then the value of $n$ is

If for the series $a_1, a_2, a_3, \ldots$,the expression $a_r - a_{r+1}$ bears a constant ratio with $a_r a_{r+1}$,then $a_1, a_2, a_3, \ldots$ are in:

The harmonic mean of $4, 8, 16$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module