Question 9 & 10 Exercise 4.3

Solutions of Question 9 & 10 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.

Find the sum 'of all multiples of 9 between 300 and 700.

All the multiples of 9 between 300 and 700 are:
$$306,315,324,333, \ldots, 693$$.
Here, $a=306$,
$$d=(315-306) = 9 \text { and } a_n=693 .$$
Let the number of terms be $n$. Then
\begin{align}a_n&=a_1+(n-1) d \text { becomes } \\ \Rightarrow a_1+(n-1) d&=693 \\ \Rightarrow 306+(n-1) \cdot 9&=693 \\ \Rightarrow 9 n&=396 \\ \Rightarrow n&=44.\end{align} $\therefore$ Required sum is:
\begin{align} S_{44}&=\dfrac{44}{2}(306+693)\\ \Rightarrow S_{44}&=22(306+693)\\ \Rightarrow S_{44}&=22(999)\\ \Rightarrow S_{44}&=21987.\end{align} Hence, sum of all multiples of $9$ lying between $300$ and $700$ is equal to $21,978$.

The sum of Rs. $1000$ is distributed among four people so that each person after the first receives Rs. $20$ less than the preceding person. How much does each person receive?

The total money for distribution $S_4=1000$,
therefore we have $n=4$
\begin{align}\text{Let the first person receives}&=R s . a\\ \text{Then the seçond receives}&=\operatorname{Rs}(a-20)\\ \text{The third receives}&=R s .(a-20-20)=R s .(a-40)\\ \text{the fourth reccives}&=R s .(a-40-20)=R s .(a-60) \text {. }\end{align} hence the $a, a-20, a-40, a-60$ forms an arithmetic sequence with $d=-20$, we know
\begin{align}S_n&=\dfrac{n}{2}[2 a+(n-1) d] \\ \therefore S_4&=\dfrac{4}{2}[2 a+3(-20)] . \\ \Rightarrow 2 a-60&=\dfrac{1000}{2}=500 \\ \Rightarrow 2 a&=560 \text { or } a=\dfrac{560}{2}=280\\ \text{The first person reecives}&=R \delta . a=R s .280\\ \text{The second person receives}&=\text { Rs. }(a-20)=R s \cdot 260 \text {. }\\ \text{The third person receives}&=R s \cdot(a-40)=R s .240\\ \text{the fourth person receives} &=R s .(a-60)=R s .220\\ \text{Rs}280,\text{RS}260,&\text{RS}240,\text{Rs}220\end{align}