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What is the sum of all two-digit odd positive numbers?
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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,
$\mathrm{S}_{\mathrm{N}}=\left(\frac{\mathrm{N}}{2}\right)\left[2 \mathrm{~T}_{1}+(\mathrm{N}-1) \mathrm{d}\right.$
$\Rightarrow \mathrm{S}_{\mathrm{N}}=\left(\frac{45}{2}\right)[2 \times 11+(45-1)(2)]$
$\Rightarrow \mathrm{S}_{\mathrm{N}}=\left(\frac{45}{2}\right)[22+88]$
$\Rightarrow S_{N}=\left(\frac{45}{2}\right)[110]$
$\Rightarrow \mathrm{S}_{\mathrm{N}}=45 \times 55$
$\Rightarrow S_{N}=2475$

by Diamond (75,914 points)

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