**Answer**

2475

**Explanation:**

We know that smallest two digit odd number is 11, and the greatest two digit number is 99.

Since difference between consecutive odd numbers is 2 , these numbers form an arithmetic progression with first term $=11$, last term $=99$ and difference $=2$.

If there are total $\mathrm{N}$ terms in series, $\mathrm{N}^{\text {th }}$ term is given by $\mathrm{T}_{\mathrm{N}}=\mathrm{T}_{1}+(\mathrm{N}-1) \mathrm{d}$

$\Rightarrow 99=11+(\mathrm{N}-1)(2)$

$\Rightarrow 2(\mathrm{~N}-1)=99-11$

$2(\mathrm{~N}-1)=88$

$\Rightarrow \mathrm{N}-1=\frac{88}{2}$

$\Rightarrow \mathrm{N}=44+1$

$\Rightarrow \mathrm{N}=45$

Now sum of arithmetic progression can be found using standard formula,

$\mathrm{S}_{\mathrm{N}}=\left(\frac{\mathrm{N}}{2}\right)\left[2 \mathrm{~T}_{1}+(\mathrm{N}-1) \mathrm{d}\right.$

$\Rightarrow \mathrm{S}_{\mathrm{N}}=\left(\frac{45}{2}\right)[2 \times 11+(45-1)(2)]$

$\Rightarrow \mathrm{S}_{\mathrm{N}}=\left(\frac{45}{2}\right)[22+88]$

$\Rightarrow S_{N}=\left(\frac{45}{2}\right)[110]$

$\Rightarrow \mathrm{S}_{\mathrm{N}}=45 \times 55$

$\Rightarrow S_{N}=2475$