The sum of first 50 positive even integers is 2,550

[Guest]

Source: Gmatprep 2

The sum of first 50 positive even integers is 2,550. what is the sum of even integers from 102 to 200, inclusive

a) 5100
b) 7550
c)10,100
d) 15,500
e) 20,100

OA is B. Can someone explain how to do this problem by smart way?

[JPG]

Source: Gmatprep 2

The sum of first 50 positive even integers is 2,550. what is the sum of even integers from 102 to 200, inclusive

a) 5100
b) 7550
c)10,100
d) 15,500
e) 20,100

I don't think this is "smart" but it is one way to tackle it. The first thing to recognize is that they are asking about consecutive integers. Anytime you want to find the sum of a set of consecutive integers you look for the mean or median of the set. For a set of consecutive integers, the mean or median will always be the middle term.

The first sentence helps us to realize that the second set has a total of 50 terms. The reason is that the average of the first set is 50 (2550 / 50) and 50 must be the middle number of 50 terms so the set is from 2 - 100. Thus the second set will also have 50 terms and the middle term will be 151 (avg. of 150 and 152).
151 * 50 = 7550

[Guest]

The way i went to solve this question is:

Given: 2+4+6+8+.......+100 = 2,550

Find: 102+104+106+......+200?

Steps: 102+104+106+......+200
(100+2)+(100+4)+(100+6)+.....+(100+100)
10*100 + (2+4+6+8+....+100)
1000+2550
3,550

CAn someone explain what is wrong in the above calculation?

[Guest]

Oh...Ma Bad!

I figured out!!! It shud be 50*100 +2550 =7550

[RonPurewal]

http://www.manhattangmat.com/forums/sum ... t4166.html

[Saurav]

There is a generic method to solve addition problems like these.

Question asks sum of even integers from 102 to 200 inclusive

S= 102 + 104 + ... 198 + 200

This can be represented by a series where each term Tn=100 + 2n (1<= n <= 50)

S= Sum(1<= n <= 50) [Tn]
S= Sum(1<= n <= 50) [100 + 2n]
S= 50 x 100 + 2 x Sum(1<= n <= 50)n

the term Sum(1<= n <= 50)n represents the sum of integers from 1 to 50 = 50 x (50+1) / 2

S= 50 x 100 + 2 x 50 x (50+1) / 2
S= 5000 + 50 x (50+1)
S= 5000 + 50 x 51 = 7550

this can be used to find sum of all multiples of 3 less than 100, Tn = 3n where 1 <= n <= 33. The sum = 3 x (sum of integers from 1 to 33)

[RonPurewal]

There is a generic method to solve addition problems like these.

Question asks sum of even integers from 102 to 200 inclusive

S= 102 + 104 + ... 198 + 200

This can be represented by a series where each term Tn=100 + 2n (1<= n <= 50)

S= Sum(1<= n <= 50) [Tn]
S= Sum(1<= n <= 50) [100 + 2n]
S= 50 x 100 + 2 x Sum(1<= n <= 50)n

the term Sum(1<= n <= 50)n represents the sum of integers from 1 to 50 = 50 x (50+1) / 2

S= 50 x 100 + 2 x 50 x (50+1) / 2
S= 5000 + 50 x (50+1)
S= 5000 + 50 x 51 = 7550

this can be used to find sum of all multiples of 3 less than 100, Tn = 3n where 1 <= n <= 33. The sum = 3 x (sum of integers from 1 to 33)

this is all good, but make sure that you can do pattern recognition. in other words, this is a pretty formula, and it can't hurt you to know it, but the chance that you'll actually get to use it on the exam is pretty small. by contrast, the chance that you'll get to use pattern recognition skills on the actual exam is very, very high indeed.

by the way, the actual formula is somewhat inaccessible in this post. the post is certainly correct, but here's the formula in more accessible terms:
the sum of all integers from 1 to n, where n is a positive integer, is n(n + 1) / 2.
again, though, pattern recognition is king. unless you have an extremely easy time memorizing things like this, it's probably not worth the effort, especially vis-à-vis developing conceptual skills such as pattern recognition.