Question

Find the sum of all two-digit odd positive numbers.

Answer 1

Step 1: Find the number of terms in the AP:

Formula:

The $n^{\mathrm{th}}$ term of the AP is $a_n=a+\left(n-1\right)d$.

The two-digit odd positive numbers are:

$11,13,15,17,19,\dots ,99$.
Substitute $a=11$, $d=2$ and $a_n=99$ into the $n^{\mathrm{th}}$ term of the AP formula.
$$\begin{gathered} \Rightarrow 99=11+\left(n-1\right)\left(2\right) \\ \Rightarrow 99=11+2n-2 \\ \Rightarrow 2n=99-9 \\ \Rightarrow 2n=90 \\ \Rightarrow n=45 \end{gathered}$$

Step 2: Evaluate the sum of $\mathbf{45}$ terms.

Formula:

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

The sum of all two-digit odd positive numbers will be:
$\begin{array}{rcl}& \Rightarrow & S_{45}=\frac{45}{2}\left[2\left(11\right)+\left(45-1\right)2\right]\\ & =& \frac{45}{2}\times \left(22+88\right)\\ & =& \frac{45}{2}\times 110\\ & =& 45\times 55\\ & =& 2475\end{array}$

Hence, the sum of all two-digit odd positive numbers is $2475$.