What is the ratio of the sum of the even positive integers between 1

Question

What is the ratio of the sum of the even positive integers between 1 and 100 (inclusive) and the sum of the odd positive integers between 100 and 150?

(A) 102 to 125
(B) 50 to 51
(C) 51 to 56
(D) 202 to 251
(E) 2 to 3

Charli08 · Answer

Sum of an arithmetic sequence:  (First term + last term) * (Number of terms/2)

Sum of even numbers between 1 and 100  inclusive :

(2+100) * (50/2)

Sum of odd numbers between 100 and 150  inclusive :

(101 + 149) * 25/2
 
No we can do the math but this would take a long time. Instead lets simplify our equations:

102 * 50/2 =  102 * 25

250 * 25/2 =  125 * 25

We can now just cancel out the 25s as we are looking for a ratio! Hence our answer:

102 to 125

==>  A

kungfury42 · Answer

Between 1 to 100 (inclusive) there are 50 even positive integers forming an AP with first term = 2, last term = 100, and common difference = 2

Sum of this AP = 50/2*(2+100) = 50/2*102 = 50*51

Between 100 to 150 (inclusive) there are 25 odd positive integers forming an AP with first term = 101, last term = 149, and common difference = 2

Sum of this AP = 25/2*(101+149) = 25/2*250 = 25*125

Ratio of the sums = (50*51)/(25*125) = (2*51)/(125) = 102/125

Hence, option A.

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GMATinsight · Answer

the sum of the even positive integers between 1 and 100 (inclusive) = 2+4+6+....+100 (total 50 terms)  = (2+100)*\frac{50}{2} = 50*51

the sum of the Odd positive integers between 100 and 150 (inclusive) = 101+103+105+....+149 (total 25 terms)  = (101+149)*\frac{25}{2} = 125*25

Required ratio  = \frac{50*51}{125*25} = \frac{102}{125}

ANswer: Option A

Adarsh_24 · Answer

Another method:

Sum of first n even = n(n+1)
Sum of first n odd= n^2

50*51/(75^2 - 50^2)
=50*51/(75+50)(75-50)
=50*51/125*25
=102/125

bumpbot · Answer

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What is the ratio of the sum of the even positive integers between 1

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What is the ratio of the sum of the even positive integers between 1

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