Question 9 & 10 Exercise 4.5

Solutions of Question 9 & 10 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.

The sum of first six terms of a geometric series is $9$ times the sum of its three terms. Find the common ratio.

We know that the sum of $n$ terms of geometric series with common ratio $r$ and first term $a_1$ is:
The sum of six terms are:
The sum of the three terms is:
$$S_3=\dfrac{a_1(r^3-1)}{r-1} \text {. }$$
The sum of the $3$ terms is $9$ times the sum of $6$ terms, therefore \begin{align} \dfrac{a_1(r^6-1)}{r-1}&=9 \dfrac{a_1(r^3-1)}{r-1} \\ \Rightarrow r^6-1-9(r^3-1) \\ \Rightarrow r^6-1&=9 r^3-9 \\ \Rightarrow r^6&=9 r^3-8 \\ \Rightarrow r^6-9 r^3+8&=0, \\ \Rightarrow r^6-r^3-8 r^3+8&=0 \\ \Rightarrow r^3(r^3-1)-8(r^3-1)&=0 \\ \Rightarrow(r^3-1)(r^3-8)&=0 \\ \Rightarrow r^3&=1 \text { or } r^3=8 \\ \Rightarrow r&=1 \text { or } r=2.\end{align} But $r$ can not be 1 thus $r=3$.

How many terms of the series: $1+\sqrt{3}+3+\ldots$ be added to get the sum $40+13 \sqrt{3}$

Here $a_1=1$ and $r=\sqrt{3}$,
We have to find $n$ such that
$$S_n=40+13 \sqrt{3} \text {. }$$
We know that:
putting the given
\begin{align}40+13 \sqrt{3}&=\dfrac{1[(\sqrt{3})^n-1]}{\sqrt{3}-1} \\ \Rightarrow(\sqrt{3})^n-1&=(40+13 \sqrt{3})(\sqrt{3}-1) \\ \Rightarrow(\sqrt{3}^n-1&=40 \sqrt{3}-40+39-13 \sqrt{3} \\ \Rightarrow(\sqrt{3})^n-1&=(40-13) \sqrt{3}-1 \\ \Rightarrow(\sqrt{3})^n&=27 \sqrt{3} \\ \Rightarrow 3 \dfrac{n}{2}&=3^3 \cdot 3 \dfrac{1}{2} \\ \Rightarrow 3 \dfrac{7}{2}&=3 \dfrac{7}{2} \\ \Rightarrow \dfrac{n}{2}&=\dfrac{7}{2} \\ \Rightarrow n&=7,\end{align} hence if $7$ term of the given sequence are added then the sum will be $40+13 \sqrt{3}$.