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If the terms of a sequence {An} are defined by An=n/100, whe

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If the terms of a sequence {An} are defined by An=n/100, whe

by Max@Math Revolution » Fri May 11, 2018 12:19 am

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[GMAT math practice question]

If the terms of a sequence {An} are defined by An=n/100, where n ranges over all the integers between 101 and 200, inclusive, what is the sum of all the An?

A. 150
B. 150.5
C. 151
D. 200
E. 201
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by GMATGuruNY » Fri May 11, 2018 4:16 am
Max@Math Revolution wrote:[GMAT math practice question]

If the terms of a sequence {An} are defined by An=n/100, where n ranges over all the integers between 101 and 200, inclusive, what is the sum of all the An?

A. 150
B. 150.5
C. 151
D. 200
E. 201
For any EVENLY SPACED SET:
Count = (biggest - smallest)/(increment) + 1.
Average = (biggest + smallest)/2.
Sum = (count)(average).
The INCREMENT is the difference between successive values.

Integers between 101 and 200, inclusive:
Here, the integers are CONSECUTIVE, so the increment = 1.
Count = (200-101)/1 + 1 = 100.
Average = (200 + 101)/2 = 301/2.
Sum = (100)(301/2) = 15050.

Since every term in sequence A is 1/100 of every term in the set above, the sum for sequence A must be 1/100 of the blue sum above:
15050/100 = 150.5.

The correct answer is B.
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by Max@Math Revolution » Sun May 13, 2018 5:35 pm
=>

A101 + A102 + ... + A200
= 101/100 + 102/100 + ... + 200/100
= (1/100)( 101 + 102 + ... + 200 )
= (1/100) ( 100*( 101 + 200 ) / 2 )
= 301/2 = 150.5

Therefore, the answer is B.
Answer: B
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by Jeff@TargetTestPrep » Tue May 15, 2018 3:44 pm
Max@Math Revolution wrote:[GMAT math practice question]

If the terms of a sequence {An} are defined by An=n/100, where n ranges over all the integers between 101 and 200, inclusive, what is the sum of all the An?

A. 150
B. 150.5
C. 151
D. 200
E. 201
We see that the terms are: 101/100, 102/100, 103/100, ..., 200/100. So the sum is:

101/100 + 102/100 + 103/100 + ... + 200/100

(101 + 102 + 103 + ... + 200)/100

Notice that the numerator is the sum of all the integers from 101 to 200, inclusive. Since there are 100 integers in this sum and the average is (101 + 200)/2 = 150.5, then the sum (or the numerator) is 100 x 150.5, and we have:

(100 x 150.5)/100

150.5

Answer: B

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by swerve » Wed May 16, 2018 8:20 am
An=101/100 + 102/100 + ........................................... + 200/100

= 1/100 * (101+102+103+.....+200)

= 1/100 * 100/2 * (101+200)

= 301/2

= 150.5 Option B.

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