Question 12 Exercise 4.4
Solutions of Question 12 of Exercise 4.4 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 12
For what value of $n, . \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is geometric mean between $a$ and $b$,
where $a$ and $b$ are not zero simultaneously.
Solution
Let $\dfrac{a^{n+1}+b^{n-1}}{a^n+b^n}$ be geometric mean between $a$ and $b$, then
\begin{align}\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}&=\sqrt{a b}\quad \because G \cdot M=\sqrt{a b} \\
\Rightarrow \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}&=a^{\dfrac{1}{2}} b^{\dfrac{1}{2}} \\
\Rightarrow a^{n+1}+b^{n+1}&=(a^n+b^n) a^{\dfrac{1}{2}} b^ \dfrac{1}{2} \\
\Rightarrow a^{n+\dfrac{1}{2}+\dfrac{1}{2}}+b^{n+\dfrac{1}{2}+\dfrac{1}{2}}
&= a^{n+\dfrac{1}{2}} b^{\dfrac{1}{2}}+a^{\dfrac{1}{2}} b^{n+\dfrac{1}{2}} \\
\Rightarrow a^{\dfrac{1}{2}} a^{n+\dfrac{1}{2}}+b^{\dfrac{1}{2}} b^{n+\dfrac{1}{2}}&=a^{n+\dfrac{1}{2}} b^{\dfrac{1}{2}}+a^{\dfrac{1}{2}} b^{n+\dfrac{1}{2}} \\
\Rightarrow a^{\dfrac{1}{2}} a^{n+\dfrac{1}{2}}-a^{n+\dfrac{1}{2}} b^{\dfrac{1}{2}}& =a^{\dfrac{1}{2}} b^{n+\dfrac{1}{2}}-b^{\dfrac{1}{2}} b^{n+\dfrac{1}{2}} \\
\Rightarrow a^{n+\dfrac{1}{2}}(a^{\dfrac{1}{2}}-b^{\dfrac{1}{2}})& =b^{n+\dfrac{1}{2}}(a^{\dfrac{1}{2}}-b^{\dfrac{1}{2}}) \\
\Rightarrow a^{n+\dfrac{1}{2}}&=b^{n+\dfrac{1}{2}} \\
(\dfrac{a}{b})^{n+\dfrac{1}{2}}&=(\dfrac{a}{b})^0 \\
\Rightarrow n+\dfrac{1}{2}&=0\\
n&=-\dfrac{1}{2}\end{align}
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