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Require sum of 1+3+5+7+9....99
Applying formula for summation of n digits with a as first digit and d is the difference
sum =Â $$\frac{n}{2} (2a + (n-1) d)$$
or this formula can be reduced to $$\frac{n}{2} \frac{a + l}{2} $$hence for calculating avg. it will be
$$\frac{a + l}{2}$$ (where $$l$$ is last term)
so $$\frac{1 + 99}{2}$$ = 50
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