Question 9 Exercise 4.4

Solutions of Question 9 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.

Insert five geometric means between $3 \dfrac{5}{9}=\dfrac{32}{9}\quad$ and $\quad40 \dfrac{1}{2}=\dfrac{81}{2}$.

Let $G_1, G_2, G_3, G_4$ and $G_5$ be the five geometric means between $\dfrac{32}{9}$ and $\dfrac{81}{2}$,
then $\dfrac{32}{9}, G_1, G_2, G_3, G_4, G_5, \dfrac{81}{2}$ forms geometric sequence of 7 terms, with $a_7=\dfrac{81}{2}$ and $a_1=\dfrac{32}{9}$.
Therefore, $$\begin{align}a_1 r^6&=\dfrac{81}{2} \\ \Rightarrow \dfrac{32}{9} r^6&=\dfrac{81}{2}\\ \Rightarrow r^6&=\dfrac{81}{2} \times \dfrac{9}{32}=\dfrac{3^4}{2} \times \dfrac{3^2}{2^5}\\ \Rightarrow r^6&=(\dfrac{3}{2})^6\\ \Rightarrow r&=\dfrac{3}{2}.\end{align}$$
$$\begin{align}G_1&=a_1 r=\dfrac{32}{9} \times \dfrac{3}{2}=\dfrac{16}{3} \\ G_2&=a_1 r^2=\dfrac{32}{9} \times \dfrac{9}{4}=8 \\ G_3&=a_1 r^3=\dfrac{32}{9} \times \dfrac{27}{8}=12 \\ G_4&=a_1 r^4=\dfrac{32}{9} \times \dfrac{81}{16}=18 \\ G_5&=a_1 r^5=\dfrac{32}{9} \times \dfrac{243}{32}=27\\ &\dfrac{16}{3},8,12,18,27\end{align}$$

Insert $6$ geometric means between $14$ and $-\dfrac{7}{64}$.

Let $G_1, G_2, G_3, G_4, G_5$ and $G_6$ be the six geometric means between $14$ and $-\dfrac{7}{64}$,
then $14, G_1, G_2, G_3, G_4, G_5, G_6,-\dfrac{7}{64}$ forms geometric sequence of $8$ terms, with
$$\begin{align}a_8&=-\dfrac{7}{64} \text { and }\\ a_1&=14 . \text { Therefore, } \\ a_1 r^7&=-\dfrac{7}{64} \\ \Rightarrow 14 r^7&=-\dfrac{7}{64} \\ \Rightarrow \quad r^7&=-\dfrac{7}{64} \cdot \times \dfrac{1}{14}=(-\dfrac{1}{2})^7 \\ \Rightarrow r&=-\dfrac{1}{2} . \\ G_1&=a_1 r=14 \times-\dfrac{1}{2}=-7 \\ G_2&=a_1 r^2=14 \times \dfrac{1}{4}=\dfrac{7}{2} \\ G_3&=a_1 r^3=14 \times-\dfrac{1}{8}=-\dfrac{7}{4} \\ G_4&=a_1 r^4=14 \times \dfrac{1}{16}=\dfrac{7}{8} \\ G_5&=a_1 r^5=14 \times-\dfrac{1}{32}=-\dfrac{7}{16} \\ G_6&=a_1 r^6=14 \times \dfrac{1}{64}=\dfrac{7}{32}\\ -7,\dfrac{7}{2},\dfrac{7}{4}&,\dfrac{7}{8},\dfrac{7}{16},\dfrac{7}{32}\end{align}$$

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