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

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What is the sum of the integers 45 through 175 inclusive? [#permalink]

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New post 22 Dec 2015, 04:26
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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
Spoiler: OA

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Re: What is the sum of the integers 45 through 175 inclusive? [#permalink]

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New post 22 Dec 2015, 04:37
D is the answer

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

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

\(131*110= D\)
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What is the sum of the integers 45 through 175 inclusive? [#permalink]

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New post 22 Dec 2015, 04:50
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Bunuel wrote:
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



Two ways to do it..

algebric way..
sum of n consecutive positive integers=\(\frac{n(n+1)}{2}\)..
in one case n=44 and other 175... subtract the sum to get the answer
sum of first 175 +ive numbers=\(175*\frac{176}{2}=15400\)..
sum of first 44 +ive numbers=\(44*\frac{45}{2}=990\)..
answer=\(15400-990=14410\)..

average mean method..
since they are consecutive number we can use average method too...
average of all numbers from 45 to 175 = average of 45 and 175=\(\frac{45+175}{2}=110\)
total integers = \(175-45+1=131...\)
sum =\(131*110=14410..\) ans
D
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Re: What is the sum of the integers 45 through 175 inclusive? [#permalink]

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New post 03 Jan 2017, 11:30
Bunuel wrote:
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


45, 46 , 47....173,174,175

a = 45
n = 175 - 45 + 1 => 131
d = 1

\(s = \frac{n}{2}[2a + (n - 1)d]\)

\(s = \frac{131}{2}[2*45 + (131 - 1)1]\)

\(s = \frac{131}{2}[90 + 130]\)

\(s = \frac{131}{2}[220]\)

\(s = 131*110\)

\(s = 14410\)

Hence, answer will be (D) 14,410
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Re: What is the sum of the integers 45 through 175 inclusive? [#permalink]

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New post 03 Nov 2017, 20:18
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
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What is the sum of the integers 45 through 175 inclusive? [#permalink]

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New post 04 Nov 2017, 11:33
Bunuel wrote:
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

(# of terms)(average) = Sum

1) Number of terms:\((\frac{LastTerm-FirstTerm}{Increment} + 1)\)

\((\frac{175-45}{1}+1)=130+1 = 131\) terms*

2) Average (arith. mean): \(\frac{First+ Last}{2}\)

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

3) (# of terms)(average) = Sum
(131)(110) =
(130 + 1)(100 + 10), FOIL
13,000 + 1,300 + 100 + 10 = 14,410

Answer D

*This formula is the general one -- can be used for multiples, odds, evens, consecutive integer sequences that do not begin with 1, etc. The "increment" part in questions about consecutive integers, which differ by 1, often is omitted
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What is the sum of the integers 45 through 175 inclusive? [#permalink]

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New post 04 Nov 2017, 18:08
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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What is the sum of the integers 45 through 175 inclusive?   [#permalink] 04 Nov 2017, 18:08
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