====== 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.
=====Question 11=====
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?
====Solution====
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 $576 \mathrm{ft}$.\\
=====Question 12=====
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?
====Solution====
\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.
====Go To====
[[math-11-kpk:sol:unit04:ex4-3-p6 |< Question 9 & 10 ]]
[[math-11-kpk:sol:unit04:ex4-3-p8|Question 13 & 14 >]]