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

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(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

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$-1/2$

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(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$.

List-$I$	List-$II$
$1$. Nitrogen molecules	$(A)$ Continuous spectrum
$2$. Incandescent solids	$(B)$ Absorption spectrum
$3$. Fraunhoffer lines	$(C)$ Band spectrum
$4$. Electric arc between iron rods	$(D)$ Emission spectrum

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

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