If the geometric mean between a and b is frac | vedclass

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

Question

(B) Given that the geometric mean between $a$ and $b$ is $\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = (ab)^{1/2}$.Multiplying both sides by $(a^n + b^n)$,we

Vedclass · Accepted Answer

$-1/2$

Vedclass · Answer

(B) Given that the geometric mean between $a$ and $b$ is $\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = (ab)^{1/2}$.Multiplying both sides by $(a^n + b^n)$,we get:$a^{n+1} + b^{n+1} = (ab)^{1/2}(a^n + b^n)$$a^{n+1} + b^{n+1} = a^{n+1/2}b^{1/2} + a^{1/2}b^{n+1/2}$Rearranging the terms:$a^{n+1} - a^{n+1/2}b^{1/2} + b^{n+1} - a^{1/2}b^{n+1/2} = 0$$a^{n+1/2}(a^{1/2} - b^{1/2}) - b^{n+1/2}(a^{1/2} - b^{1/2}) = 0$$(a^{n+1/2} - b^{n+1/2})(a^{1/2} - b^{1/2}) = 0$Since $a 
eq b$,we have $a^{1/2} 
eq b^{1/2}$,so:$a^{n+1/2} - b^{n+1/2} = 0$$a^{n+1/2} = b^{n+1/2}$$(\frac{a}{b})^{n+1/2} = 1 = (\frac{a}{b})^0$Equating the exponents:$n + 1/2 = 0$$n = -1/2$.

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

Similar Questions

Assertion : $LiCl$ is predominantly a covalent compound.
Reason : Electronegativity difference between $Li$ and $Cl$ is too small.

Which stage of prophase-$I$ is represented in the given figure?

The standard deviation of the numbers $22, 26, 28, 20, 24, 30$ is

$P, Q, R$ and $S$ are four points with the position vectors $3i - 4j + 5k, 2i - j + 4k, -4i + 5j + k$ and $-3i + 4j + 3k$, respectively. Then, the line $PQ$ meets the line $RS$ at the point

$A$ particle at the end of a spring executes simple harmonic motion with a period $t_1$,while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $T$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module