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Consecutive Integers

This topic has 5 expert replies and 4 member replies
NeilWatson Senior | Next Rank: 100 Posts
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Consecutive Integers

Post Fri Apr 11, 2014 7:27 pm
For any positive integer n, the sum of the first n integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301. Would like a very thorough explanation of this problem.

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Gurpreet singh Senior | Next Rank: 100 Posts Default Avatar
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Post Fri May 27, 2016 6:03 pm
Thank you. very structured explanation.

gmat_chanakya wrote:
The main question here is whats the sum of all even integers between 99 and 301.

1. The set we are interested is {100, 102, 104, ...., 300}
2. Average = Sum/number of items and this implies Sum = Average * Number of Items
3. Number of Items for a evenly Spaced Set = [(Last - First)/Increment]+1 = [(300-100)/2]+1 = 101
4. Average for evenly spaced set = Median of the Set = (First+Last)/2 = (100+300)/2 = 200
5. Plug it back in 3. Sum = 101*200 = 202,00

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shwetagoyal121 Newbie | Next Rank: 10 Posts
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Post Mon May 30, 2016 3:43 am
NeilWatson wrote:
For any positive integer n, the sum of the first n integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301. Would like a very thorough explanation of this problem.
You don't need to follow the formula. Just go by the simple method, i.e,

sum = Average* no. of terms

Average of the series = average of any two corresponding numbers (eg. the first and last term)

(100+300)/2 = 200

Number of terms = (300-100)/2 + 1 = 101

Hence, sum= 101*200 = 20200

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danielle07 Senior | Next Rank: 100 Posts Default Avatar
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Post Sat Sep 02, 2017 2:00 am
What to do first in this kind of problem?

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gmat_chanakya Junior | Next Rank: 30 Posts Default Avatar
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Post Mon Apr 14, 2014 3:50 pm
The main question here is whats the sum of all even integers between 99 and 301.

1. The set we are interested is {100, 102, 104, ...., 300}
2. Average = Sum/number of items and this implies Sum = Average * Number of Items
3. Number of Items for a evenly Spaced Set = [(Last - First)/Increment]+1 = [(300-100)/2]+1 = 101
4. Average for evenly spaced set = Median of the Set = (First+Last)/2 = (100+300)/2 = 200
5. Plug it back in 3. Sum = 101*200 = 202,00

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Matt@VeritasPrep GMAT Instructor
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Post Sat Apr 12, 2014 1:29 pm
An important method to take away from this problem is that ANY evenly spaced set can be added in this fashion:

(Sum of an evenly spaced set) = (Average of the least and greatest members of the set) * (# of #'s in the set)

This works for the reasons that Brent described above: since every pair of numbers equidistant from the median of the set adds up to DOUBLE the median of the set, for addition purposes we're essentially just adding the median to itself again and again.

Here's an example. Take the set {2, 4, 6, 8, 10}. Notice that 2 + 10 = 6 + 6 and 4 + 8 = 6 + 6. So the sum of the set is really just 6+6+6+6+6, or 6*5, or (Median of the Set) * (# of #'s in the set).

Now take any evenly spaced set, such as {a, a + d, a + 2d, a + 3d, a + 4d}. Notice again that a + (a+4d) = (a + 2d) + (a + 2d) and that (a + d) + (a + 3d) = (a + 2d) + (a + 2d). So any pair of terms equidistant from the median adds up to double the median, and the sum of the set is (a + 2d) * 5, or (Median) * (# of #'s).

Finally, let's apply it to this problem. The median of the set is (100+300)/2, or 200, or the average of the least and greatest terms in the set. Our even integers range from 100, or 50*2, to 300, or 150*2, so we have 101 integers (from the 50th even to the 150th even). Hence our sum is 200 * 101, or 20,200.

The great thing about these sets is that the median is easy to find: it's just the average of the two extreme terms in the set, the least and the greatest!

Once you get the hang of this, ANY summation problem on the GMAT should be a piece of cake.

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Post Sat Apr 12, 2014 3:09 am
Quote:
For any positive integer n, the sum of the first n positive integers 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
Alternate approach:

100, 102, 104......198, 200, 202.....296, 298, 300.

The median is the value halfway between 100 and 300:
200.
Of the 100 integers between 100 and 199, inclusive, half are odd, half are even.
Thus, in the set above, there are 50 even integers to the LEFT of 200, implying that there are also 50 even integers to the RIGHT of 200.

Add successive PAIRS of integers, working from the OUTSIDE IN:
100+300 = 400.
102+298 = 400.
104+296 = 400.

Notice that the sum of each pair = 400.
Since there will be 50 of these pairs, the following sum is yielded:
50*400 = 20,000.
Adding in the median of 200, the final sum = 20,000 + 200 = 20,200.

The correct answer is B.

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Post Fri Apr 11, 2014 8:13 pm
Quote:
For any positive integer n, the sum of the first n positive integers 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
Approach #3:
Take 100+102+104+ ...+298+300 and factor out the 2 to get 2(50+51+52+...+149+150)
From here, we'll evaluate the sum 50+51+52+...+149+150, and then double it.

Important: notice that 50+51+.....149+150 = (sum of 1 to 150) - (sum of 1 to 49)

Now we use the given formula:
sum of 1 to 150 = 150(151)/2 = 11,325
sum of 1 to 49 = 49(50)/2 = 1,225

So, sum of 50 to 150 = 11,325 - 1,225 = 10,100

So, 2(50+51+52+...+149+150) = 2(10,100) = 20,200 = B

Cheers,
Brent

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Post Fri Apr 11, 2014 8:13 pm
Quote:
For any positive integer n, the sum of the first n positive integers 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
Approach #2:

From my last post, we can see that we have 101 even integers from 100 to 300 inclusive.

Since the values in the set are equally spaced, the average (mean) of the 101 numbers = (first number + last number)/2 = (100 + 300)/2 = 400/2 = 200

So, we have 101 integers, whose average value is 200.
So, the sum of all 101 integers = (101)(200)
= 20,200
= B

Cheers,
Brent

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Post Fri Apr 11, 2014 8:12 pm
Quote:
For any positive integer n, the sum of the first n positive integers 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
Approach #1

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150-50+1=101)

To evaluate 2(50+51+52+...+149+150) I'll add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101x200 = 20,200 = B

Cheers,
Brent

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