For any positive integer n, the sum of the first n positive integer equals n( n+ 1)/2. What is the sum
of all the even integers between 99 and 301 ?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
B
Please explain.
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quite a few way: the numbers are: 100 102 104 .... 300 since the numbers are equally spaced you may find the mean and multiply it by total numbers.farooq wrote:For any positive integer n, the sum of the first n( n+ 1)/n positive integers equals 2 . What is the sum
of all the even integers between 99 and 301 ?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
B
Please explain.
If you know the AM formula : Tn = a + (n-1)*d here 300 = 100 + (n-1)*2 or n = (300-100)/2 + 1 = 101
Sum : n/2{first term + last term} = (101/2)*400 = 20200
Using the above formula : need to convert these number from 1..n
S = 100 + 102 + 104 + ... + 300
= 100 + 100+2 + 100+4 + ...+ 100+200
= 100*101 + (2+4+6+..+200)
= 10100 + 2*(1+2+3+..100)
= 10100 + 2*100*101/2
= 20200
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Would someone please provide a concise breakdown of this problem. I am having a hard time understanding. Based on the question stem I don't get any of the answers, nor do I follow the explanations listed here or especially in the back of the OG.
My process.... N= 100 ----- (300-100)/2
100(100+1) = 10100/2 = 5050
My process.... N= 100 ----- (300-100)/2
100(100+1) = 10100/2 = 5050
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jk2010 wrote:Would someone please provide a concise breakdown of this problem. I am having a hard time understanding. Based on the question stem I don't get any of the answers, nor do I follow the explanations listed here or especially in the back of the OG.
My process.... N= 100 ----- (300-100)/2
100(100+1) = 10100/2 = 5050
Sum of evenly spaced integers is given by - no of terms * Mean of the terms
In the problem above, first even integer is 100 and last even integer is 300
no of terms = 300-100)/2 + 1 = 101.
Mean = (last term + first term)/difference between consecutive term
= 300+100/2 = 200
hence Sum = 101 * 200 = 20200
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We are given that 1 + 2 + 3 + ... + n-1 + n = n(n+1)/2 and asked to find 100 + 102 + ... + 298 + 300farooq wrote:For any positive integer n, the sum of the first n positive integer equals n( n+ 1)/2. What is the sum
of all the even integers between 99 and 301 ?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
B
Please explain.
Compare the two expressions and you will notice that we can write the second as 2(50 + 51 + ... + 149 + 150 ). Why do we do this? Simply to make it look more like the first expression so that we can use the formula provided.
Of course, we would like to start the sum with 1, so we could write
2(50 + 51 + ... + 149 + 150) = 2(1 + 2 + 3 + ... + 150) - 2( 1 + 2 + ... 49)
=2 (150)(151)/2 - 2(50)(51)/2 = 50((3(151) -51)= 50(453 - 51) = 50(402) = 20100
Do read the alternative solutions, as they are faster if you know the technique
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The question asks for the sum of the even integers from 100 to 300, inclusive.jk2010 wrote:Would someone please provide a concise breakdown of this problem. I am having a hard time understanding. Based on the question stem I don't get any of the answers, nor do I follow the explanations listed here or especially in the back of the OG.
My process.... N= 100 ----- (300-100)/2
100(100+1) = 10100/2 = 5050
Ignore the formula given. Instead, use the following:
Sum = (number of integers) * (average of biggest and smallest)
To count the number of evenly spaced integers in a set:
Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1
Since we're adding only the even integers, the distance between each successive pair is 2.
Thus, the number of even integers from 100 to 300 = (300-100)/2 + 1 = 101.
Average of biggest and smallest = (300+100)/2 = 200.
Sum = (number of integers) * (average of biggest and smallest) = 101*200 = 20,200.
The correct answer is B.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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