Question 15 Exercise 4.2

Solutions of Question 15 of Exercise 4.2 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

For what value of $n, \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is the arithmetic mean between $a$ and $b$. Where $a$ and $b$ are not zero simultaneously.

Suppose $A$ represents the arithmetic mean between $a$ and $b$, then $$A=\dfrac{a+b}{2}. --- (1)$$ Also, we have given $$A=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}. --- (2)$$ Comparing (1) and (2), we have $$\begin{align}&\dfrac{a+b}{2}=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}, --- (3) \\ \implies &(a^n+b^n)(a+b)=2(a^{n+1}+b^{n+1}) \\ \implies &a^{n+1}+a b^n+a^n b+b^{n+1}=2 a^{n+1}+2 b^{n+1} \\ \implies &a b^n+b^{n+1}-2 b^{n+1}=2 a^{n+1}-a^{n+1}-a^n b \\ \implies &a b^n-b^{n+1}=a^{n+1}-a^n b \\ \implies & b^n(a-b)=a^n(a-b)\end{align}$$ If $a\neq b$, then we have $$\begin{align} &b^n =a^n \\ \implies &\dfrac{b^n}{a^n}=1\\ \implies &\left(\dfrac{b}{a}\right)^n=\left(\dfrac{b}{a}\right)^0\\ \implies &n=0.\end{align}$$ If $a=b$, then from (3), we have $$\begin{align}&\dfrac{a+a}{2}=\dfrac{a^{n+1}+a^{n+1}}{a^n+a^n} \\ \implies &a=\dfrac{2a^{n+1}}{2a^n} \\ \implies &a=a \end{align}$$ Hence $n=0$, when $a\neq b$ and for $a=b$, given expression is A.M for all $n$.

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