Question 15 & 16 Exercise 4.5

Solutions of Question 15 & 16 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.

A man wishes to save money by setting aside Rs. I the first day, Rs. $2$ the second day, Rs. $4$ the third day, and so on, doubling the amount each day. If this continụed, how much must be set aside on the $15^{\text {th }}$ day? What is the total amoumt saved at the end of thirty days?

Let the money saves on first day $a_1=R s .1$

Saves on second day $a_2=R s .2$

Saves on third day $a_3=R s .4$ and so on.

Hence the sequence is $1,2,4,8, \ldots$ is a geometric sequence, with $a_1=1 . \quad r=2 . \quad n=15$

We know that $a_n=a_1 r^{n-1}$.

The money he saves on $15^{1 / 2}$ day is $$a_{15}=a_1 r^{14}$$ becomes in the given case $$a_{15}=1 .(2)^{1 4}=R s .16384$$ Now the total amount saved at the end of 30 days is

$$S_{30}=\dfrac{a_1(r^{30}-1)}{r-1}$$ putting $r-2$ and $a_1=1$, then

$$\begin{align}S_{30}&=\dfrac{1[2^{30}-1]}{2-1}=2^{30}-1 \\ \Rightarrow S_{30}&=R s .1073741823 \end{align}$$ $$\text{Rs.}=16384; \text{Rs.}= 1073741823$$

The number of bacteria in a culture increased geometrically from $64000$ to $729000$ in $6$ days. Find the daily rate of increase if the rate is assumed to be constant.

The number of bacteria at the start $a_1=64000$

Let number of bacteria after first day be $a_2$, after second be $a_3$ and so on.

The number of bacteria after $6$th day $a_7=729000$. Since the number is increasing geometrically,

we know that $a_n=a_1 r^{n-1}$, putting $a_1$ and $n=7$, then $$\begin{align}&729000=64000 r^{7-1} \\ & \Rightarrow r^6=\dfrac{729000}{64000} \\ & \Rightarrow r^6=\dfrac{729}{64}=\dfrac{3^6}{2^6} \\ & \Rightarrow r^6=(\dfrac{3}{2})^6 \\ & \Rightarrow r=\dfrac{3}{2}\end{align}$$ Hence the number of bacteria increasing with $\dfrac{3}{2}$ of the slarting number each day.

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