Question 11 & 12 Exercise 4.3

Solutions of Question 11 & 12 of Exercise 4.3 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 distance which an object dropped from a cliff will fall $16 \mathrm{ft}$ the first second, $48 \mathrm{ft}$ the next second, $80 \mathrm{ft}$ the third and so on. What is the total distance the object will fall in six seconds?

The distance in first second $a_1=16 \mathrm{ft}$
The distance in the $2^{\text {nd }}$ second $a_2=48 \mathrm{ft}$
The distance in the third second $a_3=80 \mathrm{ft}$ and so on.
Hence the sequence $16,48,80, \ldots \quad$ is an arithmetic sequence with $d=48-16=32$.
The total distance in six second is $S_6$ By arithmetic sum formula, we know
\begin{align}S_n&=\dfrac{n}{2}[2 a_1+(n-1) d] \\ \therefore S_6&=\dfrac{6}{2}(2.16+5.32) \\ \Rightarrow S_6&=3(192)=576 f t .\end{align} Hence the total distance covered in six seconds is $576 \mathrm{ft}$.

Afzal Khan saves Rs. $1$ the first day, Rs. $2$ the second day, Rs. 3$$ the third day and Rs.$N$ of the $n$th day for thirty days. How much does he saved at the end of the thirtieth day?

\begin{align}\text{Saves at first day}& =R s .1\\ \text{Saves at second day}&=Rs. 2\\ \text{Saves at third day}&=Rs. 3 \text{ and so on}\end{align} So the sequence formed $1,2,3,4, \ldots$ is an arithmetic sequence with $d=1$.
We have to find the total safe in thirty days means $S_{30}$.
We know that:
\begin{align}S_n&=\dfrac{n}{2}[2 a_1+(n-1) d] \\ \because S_{30}&=\dfrac{30}{2}[2.1+29.1] \\ \Rightarrow \quad S_{30}&=15(31)=465\end{align} The total money he saves in thirty days are Rs. 465.