Question

What is the sum of all odd numbers between $0\text{and}100$?

Answer 1

Determine the sum of all odd numbers between$0\text{and}100$.

Odd numbers are defined as any number which cannot be divided by two. In other words, several forms$2k+1$, where $k\in Z$ (i.e. integers) are called odd numbers.

The sum of the odd numbers are $=1+3+5+7+9+\dots .+93+95+97+99$

For any arithmetic Progression $(\mathrm{A}.\mathrm{P}.)$:

$T\left(n\right)=a+(n–1)d$

Where,

$a$ is the first time

$d$ is difference

.$n$ is the number of terms.

$\mathrm{T}\left(n\right)\mathrm{or}l$ is $n\mathrm{th}$ term

Here, the odd number between $0\text{and}100$ which $1,3,5,7,9,\dots .,93,95,97,99$ are in $\mathrm{A}.\mathrm{P}.$

where,

$$\begin{gathered} a=1 \\ l=99 \\ d=2 \end{gathered}$$

So, Calculate the no. of the term $\left(n\right)$ in $\mathrm{A}.\mathrm{P}.$:

$$\begin{gathered} \mathrm{T}\left(n\right)=a+(n–1)d \\ \Rightarrow 99=1+(n–1)2 \\ \Rightarrow 98=2n–2 \\ \Rightarrow 100=2n \\ \Rightarrow n=50 \end{gathered}$$

Calculate the sum of the terms $1+3+5+7+9+\dots .+93+95+97+99$ which is in $\mathrm{A}.\mathrm{P}.$ is as follows:

$$\begin{gathered} S\left(n\right)=\frac{n}{2}(a+l)(\because n=50) \\ \Rightarrow S\left(n\right)=\frac{50}{2}(1+99) \\ \Rightarrow S\left(n\right)=\frac{50}{2}\times 100 \\ \Rightarrow S\left(n\right)=50\times 50 \\ \Rightarrow S\left(n\right)=2500 \end{gathered}$$

Hence, the sum of all odd numbers between $0\text{and}100$ is$2500$.