This question is testing two things:

1. Find number of integers in a set

2. Find sum of integers in a set

Now,

To find the number of integers in a set (

consecutive):

Formula is

(Last term - First term) + 1= 175 - 45 + 1

= 131

Formula for sum of an arithmetic series (when there is a common

difference)

Sum = \(\frac{n}{2}\) (first term + last term) where n = number of integers

= \(\frac{131}{2}\) (45 + 175)

= \(\frac{131}{2}\) (220)

= 131 * 110

= 14,410

PS there is a trick to doing long multiplications like these. Forget about the zero for a second, and think about just 131 * 11. You can see that the digit

right before the last 0 will have to be 1. The only option that fits is D.

= 13

1 * 1

10

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