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11 May 2018, 01:28
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[GMAT math practice question]
If the terms of a sequence {An} are defined by An=\(\frac{n}{100}\), 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
If the terms of a sequence {An} are defined by An=\(\frac{n}{100}\), 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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11 May 2018, 02:07
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MathRevolution
An=\(\frac{101}{100}\) + \(\frac{102}{100}\) + ........................................... + \(\frac{200}{100}\)
= \(\frac{1}{100}\) * (101+102+103.....200)
= \(\frac{1}{100}\) * \(\frac{100}{2}\) * (101+200)
= \(\frac{301}{2}\)
= 150.5
Ans B.
11 May 2018, 16:32
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MathRevolution
The sum of an arithmetic series = (average)*(# of terms)
Average: (First Term + Last Term) ÷ 2
Number of terms: (Last Term - First Term) + 1
Methods (both are quick):
1) Use decimals
Use this sequence's numbers in decimal form. Find the sum. \(\frac{101}{100}=1.01\)
Find the sum of 1.01, 1.02 . . 1.99, 2.00
Average: \(\frac{(1.01+2.00)}{2} = \frac{3.01}{2}\) = 1.505
# of terms: (200 - 101) = 99, and (99 + 1) = 100 terms
Sum = (1.505)*(100) = 150.5
Answer B
2) forget about the fractions. Use only the integers
Find the average of the integers. Divide by 100.
Multiply that value by # of terms in the fractional sequence. Why?
*The average of the just the integers, divided by 100,
equals the average of all \(\frac{A_{n}}{100}\)
First, integers only.
Find the average of integers from 101 to 200, inclusive:
\(101 + 200 = \frac{301}{2} = 150.5\)
Second, divide that average by 100
(the integers we just averaged are all divided by 100):
\(\frac{150.5}{100} = 1.505\)
That's the average of this sequence with fractions
Third, number of terms in this sequence:
(# of terms) = (200-101= 99), 99 + 1 = 100 terms
Sum: \((1.505*100)\) \(=150.5\)
Answer B
*Try a small sample to prove the point. Say, e.g., we have this prompt's facts,
but the range is 1 to 3 inclusive. \(\frac{1}{100} = .01\)
Average:\((\frac{.01+.03}{2})=.02\)
Now try integers: average of 1 through 3 = \(\frac{1 + 3}{2} = 2\)
Divide that average by 100. \(\frac{2}{100} = .02\). That works.
