# 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.

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.

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}