Question

Find n if [a^(n+1)+b^(n+1)]/(a^n)+(b^n) may be the G. M between a and b​

Answer 1

Step-by-step explanation:

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Answer 2

ANSWER:

Given:

$\:\:\:\bullet\:\:\:\sf GM_{a\:\&\:b}=\dfrac{a^{n+1}+b^{n+1}}{(a^n)+(b^n)}$

To Find:

• Value of n

Solution:

$\text{We are given that,}\\\\:\longrightarrow\sf GM_{a\:\&\:b}=\dfrac{a^{n+1}+b^{n+1}}{(a^n)+(b^n)}\\\\\text{We know that,}\\\\:\hookrightarrow\sf GM_{x\:\&\:y}=\sqrt{xy}\\\\\text{So,}\\\\:\implies\sf GM_{a\:\&\:b}=\sqrt{ab}\\\\\text{Hence,}\\\\:\implies\sf\sqrt{ab}=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}\\\\\text{On cross-multiplying,}\\\\:\implies\left(\sqrt{ab}\right)\times\left(a^n+b^n\right)=a^{n+1}+b^{n+1}\\\\:\implies\sf\left(a^n\sqrt{ab}\right)+\left(b^n\sqrt{ab}\right)=a^{n+1}+b^{n+1}$

$\text{We know that,}\\\\:\hookrightarrow\sf\sqrt{x}=x^{\frac{1}{2}}\\\\\text{So,}\\\\:\implies\sf\left(a^n\times(ab)^{\frac{1}{2}}\right)+\left(b^n\times(ab)^{\frac{1}{2}}\right)=a^{n+1}+b^{n+1}\\\\\text{We know that,}\\\\:\hookrightarrow\sf(xy)^m=x^m\times y^m\\\\\text{So,}\\\\:\implies\sf\left(a^n\times a^{\frac{1}{2}}\times b^{\frac{1}{2}}\right)+\left(b^n\times a^{\frac{1}{2}}\times b^{\frac{1}{2}}\right)=a^{n+1}+b^{n+1}\\\\\text{We know that,}\\\\:\hookrightarrow\sf x^m\times x^n=x^{m+n}\\\\\text{So,}$

$:\implies\sf\left(a^{n+\frac{1}{2}}\times b^{\frac{1}{2}}\right)+\left(a^{\frac{1}{2}}\times b^{n+\frac{1}{2}}\right)=a^{n+1}+b^{n+1}\\\\\text{On rearranging the terms,}\\\\:\implies\sf\left(a^{n+\frac{1}{2}}\times b^{\frac{1}{2}}\right)-a^{n+1}=b^{n+1}-\left(a^{\frac{1}{2}}\times b^{n+\frac{1}{2}}\right)\\\\\text{On taking common,}\\\\:\implies\sf\left[a^{n+\frac{1}{2}}\right]\times\left[b^{\frac{1}{2}}-a^{\frac{1}{2}}\right]=\left[b^{n+\frac{1}{2}}\right]\times\left[b^{\frac{1}{2}}-a^{\frac{1}{2}}\right]$

$\text{On canceling, \left[b^{\frac{1}{2}}-a^{\frac{1}{2}}\right] on both sides,}\\\\:\implies\sf a^{n+\frac{1}{2}}=b^{n+\frac{1}{2}}\\\\\text{Transposing RHS to LHS,}\\\\:\implies\sf\dfrac{a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}=1\\\\\text{We know that,}\\\\:\hookrightarrow\sf\dfrac{x^m}{y^m}=\left(\dfrac{x}{y}\right)^m\\\\\text{So,}\\\\:\implies\sf\left(\dfrac{a}{b}\right)^{n+\frac{1}{2}}=1$

$\text{We know that,}\\\\:\hookrightarrow\sf1=x^0\\\\\text{So,}\\\\:\implies\sf\left(\dfrac{a}{b}\right)^{n+\frac{1}{2}}=\left(\dfrac{a}{b}\right)^0 \\\\\text{On comparing powers,}\\\\:\implies\sf n+\dfrac{1}{2}=0\\\\\bf{:\implies n=-\dfrac{1}{2}}\\\\\text{\bf{Value of n is \bf{-\dfrac{1}{2}} .}}$

Formulae Used:

$:\hookrightarrow\sf1)\:GM_{x\:\&\:y}=\sqrt{xy}\\\\:\hookrightarrow\sf2)\:\sqrt{x}=x^{\frac{1}{2}}\\\\:\hookrightarrow\sf3)\:(xy)^m=x^m\times y^m\\\\:\hookrightarrow\sf4)\:x^m\times x^n=x^{m+n}\\\\:\hookrightarrow\sf5)\:\dfrac{x^m}{y^m}=\left(\dfrac{x}{y}\right)^m\\\\:\hookrightarrow\sf6)\:1=x^0$

Learn More:

$\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Laws of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}$