Question

Find the sum of all two-digit odd numbers.

Answer 1

Step 1. Find the series of two digits odd numbers:

Given: A sum of all two digits odd numbers.

Odd number: A natural number that is not a multiple of $2$. i.e. $1,3,5,7,9.......$

A Series of two digits odd number is $11,13,15,17,......,99$

First two digits odd number is $11$.

Hence, $t_1=11$

The last two digits odd number is $99$.

Hence, $t_n=99$

Step 2. Find the common difference:

The formula for the common difference of an AP is $d=n^{th}term-{\left(n-1\right)}^{th}term$

consider $n=2$

Therefore,$t_n=t_2=13,t_{n-1}=t_{2-1}=t_1=11$

Substituting the value.

$$\begin{gathered} d=13-11 \\ d=2 \end{gathered}$$

Step 3. Find the number of terms in AP:

The formula $n^{th}$term of an AP is $t_n=a+\left(n-1\right)d$

Here $t_n=99,a=11,d=2$

Substituting the value

$$\begin{gathered} 99=11+(n-1)2 \\ 99=11+2n-2 \\ 99=9+2n \end{gathered}$$

Add $-9$ on both sides

$$\begin{gathered} 99-9=9+2n-9 \\ 90=2n \\ \frac{90}{2}=\frac{2n}{2} \\ 45=n \end{gathered}$$

Hence, there are $45$ two digits odd numbers.

Step 4. Find the sum of two digits odd numbers:

The formula of the sum of the first $n$ terms of an AP is $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

Where $a$ is the first term and $d$ is common difference.

Here $n=45,a=11,d=2$

Substituting the value

$$\begin{gathered} S_{45}=\frac{45}{2}\left(2\left(11\right)+\left(45-1\right)2\right) \\ {S}_{45}=22.5\left(22+\left(44\right)2\right) \\ {S}_{45}=22.5\left(22+88\right) \\ {S}_{45}=22.5\left(110\right) \\ {S}_{45}=2475 \end{gathered}$$

Hence, the sum of two digits odd numbers is $2475$.