Question 2 Exercise 4.5

Solutions of Question 2 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.

Some of the components $a_1, a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find the ones that are missing $a_1=1, \quad r=-2, \quad a_n=64$.

We first find $n$ and then $S_n$
We know $a_n=a_1 r^{n-1}$, therefore
$$\begin{align}64&=(-2)^{n-1}\\ \Rightarrow(-2)^{n-1}&=(-2)^6 \\ \Rightarrow n-1&=6 \\ \Rightarrow n&=7\\ S_7&=\dfrac{a_1[r^{\prime \prime}-1]}{r-1}\\ \text{then}\\ S_7&=\dfrac{1[(-2)^7-1]}{-2-1}\\ \Rightarrow S_7&=\dfrac{-128-1}{-3}\\ s_7&=\dfrac{129}{3}\end{align}$$

Some of the components $a_1, a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find the ones that are missing $r=\dfrac{1}{2}, a_9=1, n=9$

We first find $a_1$ and then $S_9$.
We know $$a_9=a_1 r^8$$
therefore we have
$$\begin{align}1&=a_1(\dfrac{1}{2})^{9-1}\\ &=a_1 \dfrac{1}{2^8}\\ \Rightarrow a_1&=2^8\\ S_9&=\dfrac{a_1[1-r^{\prime \prime}]}{1-r},\end{align}$$ becomes in the given case
$$\begin{align}\Rightarrow S_9&=\dfrac{2^8[1-(\dfrac{1}{2})^9]}{1-\dfrac{1}{2}} \\ \Rightarrow S_9&=2^9[1-\dfrac{1}{2^9}] \\ \Rightarrow \quad S_9&=2^9[\dfrac{2^9-1}{2^9}] \\ \Rightarrow \quad S_9&=2^9-1\\ \Rightarrow \quad S_9&=511\end{align}$$

Some of the components $a_1, a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find the ones that are missing $r=-2, S_n=-63, a_n=-96$

We know that
$$\begin{align}S_n&=\dfrac{a_1(r^{\prime \prime}-1)}{r-1}\\ &=\dfrac{a_1 r^{n-1} r-a_1}{r-1} \\ \text { or } S_n&=\dfrac{a_n r-a_1}{r-1}.\end{align}$$
becomes in the given case
$$\begin{align}-63&=\dfrac{-96(-2)-a_1}{-2-1} \\ \Rightarrow \quad 192-a_1&=189 \\ \Rightarrow \quad a_1&=192-189=3\end{align}$$
Now $$a_n=a_1 r^{n-1}$$
becomes in the given case
$$\begin{align}-96&=3 \cdot(-2)^{n-1} \\ \Rightarrow(-2)^{n-1}&=-32 \\ \Rightarrow(-2)^{n-1}&=(-2)^5 \\ \Rightarrow n-1&=5 \text { or } n=6.\end{align}$$

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