Question 9 and 10, Exercise 4.3
Solutions of Question 9 and 10 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
Question 9
Find the sum of the odd numbers from $1$ to $99$.
Solution.
Solution.
Sum of the odd numbers from $1$ to $99$ is
$$1+3+5+...+99 (50 \text{ terms}).$$
This is arithmetic series with: $a_{1}=1$, $n=50$, $d=3-1=2$.
Let $S_n$ represents sum of the arithmetic series. Then
\begin{align}
S_n&=\frac{n}{2}[2a_1+(n-1)d] \\
\implies S_{50}&=\frac{50}{2}[2(1)+(50-1)(2)]\\
&=25\times [2+98]\\
&=2500.
\end{align}
Hence the sum of the odd numbers from $1$ to $99$ is $2500$. 
Question 10
Find the sum of all multiples of 4 that are between $14$ and $523$.
Solution.
Sum of all multiples of 4 that are between $14$ and $523$.
$$16+20+24+...+520.$$
This is arithmetic series with: $a_{1}=16$, $d=20-16=4$, $a_n=520$, $n=?$.
We have
\begin{align}
& a_n=a_1+(n-1)d \\
\implies & 520=16+(n-1)(4) \\
\implies & 520=16+4n-4 \\
\implies & 4n=520-16+4 \\
\implies & 4n = 508\\ \implies & n = 127.
\end{align}
Let $S_n$ represents sum of the arithmetic series. Then
\begin{align}
S_n&=\frac{n}{2}[a_1+a_n] \\
\implies S_{127}&=\frac{127}{2}[16+520]\\
&=\frac{127}{2}\times 536\\
&=34036.
\end{align}
Hence the required sum is $34036$.
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