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The first odd three-digit number divisible by 5 is 105.

The last odd three-digit number divisible by 5 is 995.

Number of terms: \(\frac{995 - 105 }{ 10} + 1 = 90\)

Sum = \(\frac{n}{2 }[a + l]\)

=> Sum = \(\frac{90}{2} [105 + 995]\)

=> Sum = 49,500

Answer A
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Find the sum of all odd three digit numbers that are divisible by 5.

All odd three digit numbers divisible by 5 will have 5 as the units' digit:

First number = 105
Last Number = 995
Common difference, d = 10 (as alternate multiples of 5 will be odd numbers)
Number of terms, n = (Last term - First Term)/d + 1 = \(\frac{995 - 105}{10}\) + 1 = 89 + 1 = 90
Sum = n * (First Term + Last Term)/2 = 90 * \(\frac{(105 + 995)}{2}\) = 49,500

So, Answer will be A
Hope it helps!

Watch the following video to MASTER Sequence problems

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