The two digit odd positive integers are
\(11,13,15, \ldots, 99\)
\(a_{2}-a_{1}=13-11=2\)
\(a_{3}-a_{2}=15-13=2\)
\(\because a_{3}-a_{2}=a_{2}-a_{1}=2\)
Therefore, the series is in AP
Here, \(a=11, d=2\) and \(a_{n}=99\)
We know that,
$$
\begin{aligned}
&a_{n}=a+(n-1) d \\
&\Rightarrow 99=11+(n-1) 2 \\
&\Rightarrow 99-11=(n-1) 2 \\
&\Rightarrow 88=(n-1) 2 \\
&\Rightarrow 44=(n-1) \\
&\Rightarrow n=45
\end{aligned}
$$
Now, we have to find the sum of this AP
$$
\begin{aligned}
&\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\
&\Rightarrow \mathrm{S}_{45}=\frac{45}{2}[2 \times 11+(45-1) 2] \\
&\Rightarrow \mathrm{S}_{45}=45[11+44] \\
&\Rightarrow \mathrm{S}_{45}=45[55] \\
&\Rightarrow \mathrm{S}_{45}=2475
\end{aligned}
$$
Hence, the sum of all two digit odd numbers are \(2475 .\)
