# Question 7 & 8 Exercise 4.5

Solutions of Question 7 & 8 of Exercise 4.5 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.

Find the $\operatorname{sum} S_n$ of the first $n$ terms of the sequence $\{(\dfrac{1}{2})^n\}$.

The sequence is:
$$\{(\dfrac{1}{2})^n\}=\dfrac{1}{2}, \dfrac{1}{2^2}, \dfrac{1}{2^3}, \ldots$$
where $$a_1=\dfrac{1}{2}$$
and $$r=\dfrac{\dfrac{1}{2^2}}{\dfrac{1}{2}}=\dfrac{1}{2}$$
We know that
\begin{align}S_n&=\dfrac{a_1(1-r^n)}{1-r},\\ S_n&=\dfrac{\dfrac{1}{2}[1-(\dfrac{1}{2})^n]}{1-\dfrac{1}{2}}\\ \Rightarrow S_n&=\dfrac{\dfrac{1}{2}[1-\dfrac{1}{2^n}]}{\dfrac{1}{2}} \\ \Rightarrow S_n&=1-\dfrac{1}{2^n}\end{align}
is the required sum of $n$ terms.

The sum of three numbers in G.P is $38$, and their product is $1728$; find them.

Let the three number are $\dfrac{a}{r}, a . a r$
then by the given conditions $$\dfrac{a}{r}+a+a r=38 \ldots \ldots . .(1)$$ and
\begin{align}\dfrac{a}{r} \cdot a \cdot a r&=1728\\ \Rightarrow a^3&=1728\\ \Rightarrow \quad a&=12,\\ \text{putting}\text{in} (1)\\ \dfrac{12}{r}+12+12 r&=38\\ \Rightarrow \dfrac{1}{r}+1+r&=\dfrac{38}{12}=\dfrac{19}{6}\\ \Rightarrow \dfrac{r^2+r+1}{r}&=\dfrac{19}{6}\\ \Rightarrow 6 r^2+6 r+6&=19 r\\ \Rightarrow 6 r^2+6 r-19 r+6&=0 \\ \Rightarrow 6 r^2-13 r+6&=0 \\ \Rightarrow 6 r^2-9 r-4 r+6&=0 \\ \Rightarrow 3 r(2 r-3)-2(2 r-3)&=0 \\ \Rightarrow(3 r-2)(2 r-3)&=0 \\ \Rightarrow 3 r-2&=0 \text { or } 2 r-3=0 \\ \Rightarrow r=\dfrac{2}{3} \text { or } r=\dfrac{3}{2}\end{align} When $a=12$ and $r=\dfrac{2}{3}$ then
\begin{align}\dfrac{a}{r}&=\dfrac{12}{\dfrac{2}{3}}=18, \\ a&=12 \text { and } a r=12 \cdot \dfrac{2}{3}=8\end{align} When $a=12$ and $r=\dfrac{3}{2}$ then
\begin{align}\dfrac{a}{r}&=\dfrac{12}{\dfrac{3}{2}}=8, \quad a=12 \\ \text { and } a r&=12 \cdot \dfrac{3}{2}=18 .\end{align}