Question 4 & 5 Exercise 4.4

Solutions of Question 4 & 5 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.

How many terms are there in a geometric sequence in which the first and the last terms are 16 and $\dfrac{1}{64}$ respectively and $r=\dfrac{1}{2}$ ?

First term of the sequence $a_1=16$
Last term of the sequence $a_n=\dfrac{1}{64}$ Common ratio $r=\dfrac{1}{2}$
We have to find $n$
We know that $$a_n=a_1 r^{n-1} \quad \text{then}$$ \begin{align}\dfrac{1}{64}&=16(\dfrac{1}{2})^{n-1} \\ \Rightarrow(\dfrac{1}{2})^{n-1}&=\dfrac{1}{64 \times 16}=\dfrac{1}{1024} \\ \Rightarrow(\dfrac{1}{2})^{n-1}&=\dfrac{1}{2^{10}}=(\dfrac{1}{2})^{10} \\ \Rightarrow n-1&=10 \text { or } n=11\end{align} Hence the total number of terms are $11$ in sequence.

Find $x$ so that $x+7, x-3, x-8$ forms a three terms geometric sequence in the given order. Also give the sequence.

Since $a_1=x+7$, $a_2=x-3, a_3=x-8$ forms a three terms geometric sequence,
\begin{align}\therefore \dfrac{a_2}{a_1}&=\dfrac{a_3}{a_2} \\ \Rightarrow \dfrac{x-3}{x+7}&=\dfrac{x-8}{x-3} \\ \Rightarrow(x-3)^2&=(x-8)(x+7) \\ \Rightarrow x^2-6 x+9&=x^2-x-56 \\ \Rightarrow-6 \cdot+x&=-56-9 \\ \Rightarrow-5 x&=-65 \\ \Rightarrow x&=\dfrac{-65}{-5}=13 .\end{align} Thus the elements of the sequence are
\begin{align}x+7&=13+7=20 \\ x-3&=13+3=10 \quad\text { and } \\ x-8&=13-8=5\\ x=13;20,10,5\end{align}