====== 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.

=====Question 15=====
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.
====Solution====
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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