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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the value of $n$ so that $\large\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be the geometric mean between $a$ and $b$.

$\begin{array}{1 1} \large\frac{1}{2} \\ \large\frac{-1}{2} \\ 1\\ -1\end{array} $

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1 Answer

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Toolbox:
  • G.M. $'g'$ between $a$ and $b$ is $\sqrt {ab}$
  • $(x+y)^2=x^2+y^2+2xy$
  • $\large\frac{x^m}{y^m}=\big(\frac{x}{y}\big)^m$
Given that $\large\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is the G.M. between $a$ and $b$
We know that the G.M. between $a$ and $b$ =$\sqrt {ab}$
$\Rightarrow\: \large\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$$=\sqrt {ab}$
Squaring on both the sides we get
$\Rightarrow\:\bigg(\large\frac{a^{n+1}+b^{n+1}}{a^n+b^n}\bigg)^2$$=ab$
$\Rightarrow\:(a^{n+1}+b^{n+1})^2=ab(a^n+b^n)^2$.........(i)
$(x+y)^2=x^2+y^2+2xy$
$\Rightarrow\:(a^{n+1}+b^{n+1})^2=(a^{n+1})^2+(b^{n+1})^2+2.a^{n+1}.b^{n+1}$
$=a^{2n+2}+b^{2n+2}+2.a^{n+1}.b^{n+1}$
and
$(a^n+b^n)^2=\big[(a^n)^2+(b^n)^2+2.a^n.b^n\big]=a^{2n}+b^{2n}+2.a^n.b^n$
Step 2
Substituting the values in (i)
$a^{2n+2}+b^{2n+2}+2.a^{n+1}.b^{n+1}=ab(a^{2n}+b^{2n}+2.a^n.b^n)$
$\Rightarrow\:a^{2n+2}+b^{2n+2}+2.a^{n+1}.b^{n+1}=ba^{2n+1}+ab^{2n+1}+2.a^{n+1}.b^{n+1}$
$\Rightarrow\:a^{2n+2}+b^{2n+2}=ba^{2n+1}+ab^{2n+1}$
$\Rightarrow\:a^{2n+2}+b^{2n+2}-ba^{2n+1}-ab^{2n+1}=0$
$\Rightarrow\:(a^{2n+2}-ba^{2n+1})+(b^{2n+2}-ab^{2n+1})=0$
$\Rightarrow\:a^{2n+1}(a-b)+b^{2n+1}(b-a)=0$
$\Rightarrow\:a^{2n+1}(a-b)-b^{2n+1}(a-b)=0$
Taking $(a-b)$ common we get
$(a-b)(a^{2n+1}-b^{2n+1})=0$
We know that if $xy=0$ then either $x=0$ or $y=0$.
$\Rightarrow\:(a-b)=0$ or $(a^{2n+1}-b^{2n+1})=0$
But since $a\neq b$, $a^{2n+1}-b^{2n+1}=0$
$\Rightarrow\:a^{2n+1}=b^{2n+1}$
$\Rightarrow\:\large\frac{a^{2n+1}}{b^{2n+1}}$$=1$
$\Rightarrow\:\bigg(\large\frac{a}{b}\bigg)^{2n+1}$$=1$
We know that for any non zero number $x$, $x^0=1$
$\Rightarrow\:2n+1=0$ or $n=-\large\frac{1}{2}$
answered Mar 4, 2014 by rvidyagovindarajan_1
 

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