What is the sum of the integers 45 through 175 inclusive?

A. 12,295
B. 13,000
C. 14,300
D. 14,410
E. 28,820
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D is the answer

\(175+45\)=\(\frac{220}{2}\)\(=110\)

\(175-45=130+1=131\)

\(131*110= D\)
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For those who like formulas more, there's a way you can apply geometry to this. Since we're looking for the sum of everything between two numbers, you can imagine it like a trapezoid, with the bounds (45 and 175) acting as b1 and b2, and the range (175-45, plus 1) as the height.

The formula for a trapezoid is
= (1/2) * (b1 + b2) * (h)
= (1/2) * (45 + 175) * (175 - 45 + 1)
= (1/2) * (220) * (131)
= (110) * (131)

and to make it easier to write it all out

= (10) * (11) * (131)
= 14410

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.
= 131 * 110
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