Question 2 & 3 Exercise 4.4

Solutions of Question 2 & 3 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.

Suppose that the third term of a geometric sequence is $27$ and the fifth term is $243$. Find the first term and common ratio of the sequence.

Here $$a_3=27 \quad\text{and}\quad a_5=243$$ and we know
$$\begin{align}a_3&=a_1 r^2=27\\ a_5&=a_1 r^4=243.\end{align}$$ Dividing (ii) by (i), we get
\begin{align}\dfrac{a_1 r^4}{a_1 r^2}&=\dfrac{243}{27}=9 \\ \Rightarrow r^2&=9 \text { or } r= \pm 3 .\end{align} Putting this in (i), then
$$a_1(9)=27 \quad \text{then} \quad a_1=3$$ $$\text{Hence}\quad a_1=3, r= \pm 3$$

Find the seventh term of the geometric sequence that has $2$ and $-\sqrt{2}$ for its second and third term respectively.

$$\begin{align}\text{The second term is}=a_2&=a_1 r=2...(i)\\ \text{the third term is}=a_3&=a_1 r^2=-\sqrt{2}...(ii)\end{align}$$ Dividing (ii) by (i), we get
\begin{align}\dfrac{a_1 r^2}{a_1 r}&=-\dfrac{\sqrt{2}}{2}=-\dfrac{1}{\sqrt{2}}\\ \Rightarrow r&=-\dfrac{1}{\sqrt{2}}.\end{align} Putting in (i), then
\begin{align}\Rightarrow a_1(-\dfrac{1}{\sqrt{2}})&=2 \\ \Rightarrow a_1&=-2 \sqrt{2}\end{align} Now we know that
\begin{align}a_n&=a_1 r^{n-1}, \\ \text { putting } n&=7, \text { and } a_1, r \text { then we get } \\ a_7&=a_1 r^6=(-2 \sqrt{2})(-\dfrac{1}{\sqrt{2}})^6 . \\ \Rightarrow a_7&=(-2 \sqrt{2})[(-\dfrac{1}{\sqrt{2}})^2]^3 \\ & =(-2 \sqrt{2}) \cdot \dfrac{1}{8} \\ \Rightarrow a_7&=-2^{-\frac{3}{2}}\end{align}

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