For any positive integer n, the sum of the first n positive integers

Although a formula is provided in this problem, we can easily solve it using a different formula:

sum = (average)(quantity)

Let’s first determine the average.

In any set of numbers in an arithmetic sequence, we can determine the average using the formula:

(1st number in set + last number in set)/2

Remember, we must average the first even integer in the set and the last even integer in the set. So we have:

(100 + 300)/2 = 400/2 = 200

Next we have to determine the quantity. Once again, we include the first even integer in the set and the last even integer in the set. Thus, we are actually determining the quantity of even consecutive even integers from 100 to 300, inclusive.

Two key points to recognize:

1) Because we are determining the number of “even integers” in the set, we must divide by 2 after subtracting our quantities.

2) Because we are counting the consecutive even integers from 100 to 300, inclusive, we must “add 1” after doing the subtraction.

quantity = (300 – 100)/2 + 1

quantity = 200/2 + 1

quantity = 101

Finally, we can determine the sum.

sum = 200 x 101

sum = 20,200

Note that the reason this is easier than using the formula provided, is that the given formula would have to be applied several times since we don’t want the total of all of the first 300 numbers. We’d have to remember to subtract the sum of the first 99 and divide by two to count only the even numbers. But we’d also have to account for the fact that the first and the last number in the set are both even. So even though a formula is given, it isn’t very easy to use.

Answer is B.

In any set of consecutive integers the sum = average*number of integers

average = 200
total = 150-50 + 1 = 101

150 because 300 is the 150th even integer
50 because 100 is the 50th even integer

101*200 = 20200
Therefore there are 101 integers

Here's one approach.

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), let's 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.
101 x 200 = 20,200

Answer: B


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


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

Answer: B

Cheers,
Brent

Hi neerajeai, please note that the n(n+1)/2 shortcut formula is only applicable if the starting point is 1. Anytime you want to find the number of terms between two given numbers you should use the general formula ((first - last) / frequency) + 1. You can also multiply by the average at the end to get the sum.

In your case it is (((301-99)/2) + 1) * 200 = 102 * 200 = 20400. Answer choice C with a presumed typo in the unit digit?

Hope this helps!
-Ron

We are given the formula:


We're asked to find sum of all the even numbers between \(99\) and \(301\)....


We can factor out \(2\) from this sum:


In the brackets we have sum of numbers from \(50\) to \(150\). We can use the given formula for the sum of \(1\) to \(150\) and subtract the sum of \(1\) to \(49\) to find the total of this sum.


The final answer is .

In this question, they give you a formula to reference: the sum of the first n positive integers is n(n+1)/2

But wait, the question is about the sum of EVEN integers, not all positive integers.

We can understand the sum of even integers between 99 and 301 by somehow rewriting in terms of the expression they gave us. They were using the terminology of the first n positive integers.

OK, well sum of even integers between 99 and 301 is the same as between 100 and 300 - inclusive.

Same as the sum of the even integers up to 300 - sum of the even integers up to 98
Same as the sum of the first 150 even integers - sum of the first 49 even integers

So how do we translate this to the expression we were given? We need to translate our "even" integer expression into "positive integers" expression with our formula.

Sum of first 150 even integers = Sum of first 150 POSITIVE integers * 2
1, 2, 3, 4, 5 ===> 2, 4, 6, 8, 10 etc

Sum of first 49 even integers = Sum of first 49 POSITIVE integers * 2

2 * (n(n+1)/2) - 2 * (n(n+1)/2)

Where n=150 in the first case, and n = 49 in the second case

= 2* (150(151)/2) - 2 * (49(50)/2)
= 150*151 - 49*50
= 50 (3*151 - 49)
= 50 (453 - 49)
= 50 (404)
= 50 (400 + 4)
= 20,000 + 200
= 20,200

Sum of first n positive integers = \(\frac{n(n+1) }{ 2}\).

Even integers between 99 and 301 belong to the set { 100, 102, 104, 106….. 300}.
Therefore, sum of even integers between 99 and 301 = 100 + 102 + 104 + 106 + ………. + 300

Factoring out 2, we can write the expression as 2 (50 + 51 + 52 + ……. + 150). The term inside the bracket represents a sequence of positive integers. But, we cannot apply the given formula directly since these do not represent the FIRST n positive integers.

Therefore, (50 + 51 + 52 +…. 150) = (1+ 2 + 3 + … + 150) – (1+2+3+…. 49)

Value of the first bracket = \(\frac{150 * 151 }{ 2}\)

Value of the second bracket = \(\frac{49 * 50 }{ 2}\).

Therefore, (50 + 51 + 52 +…. 150) = \(\frac{150 * 151 }{ 2} - \frac{49 * 50 }{ 2}\) = \(\frac{50}{2}\) (3*151 – 49) = 25 (453 – 49) = 25 * 404.

This should be a small value over 20000 since 25 * 400 = 20000. The only value that fits is 20,200.

The correct answer option is B.

Hope that helps!
Aravind B T

sum of all even integer 1 to 301 = 2 * (1+2+...+150) = (2*150*151)/2 = 150*151
sum of all even integer 1 to 99 = 2 * (1+2+....+49) = (2*49*50)/2 = 49*50

required sum = 150*151 - 49*50 = 50*(453 - 49) =
404 * 50 = 20200

They are basically asking for the sum of even integers between/inclusive of 100 and 300.

Sum = \(100+102+ .... 300 = 2(50+51+52+ ....150)\) - We've converted even integers into consecutive integers for using this formula.

Now, let's just add and subtract 1 to 50 to this series, so we get:

\(2[(1+2+3 ..... 50+51+ ..... 150)-(1+2+3 ...50)]\)

So, we've broken this sum down into two sums of 1 .... x

Sum of 1 .... 150 = \(\frac{150(150+1)}{2} = 75*151\)

Similarly, sum of 1 .... 49 = \(\frac{49(49+1)}{2} = 25*49\)

So we get the answer to be: \(2[(75*151) - (25*49) = 20200\)

The reason why we do this is because in the sum of even integers, when 2 is factored out, you get consecutive integers as I've shown here. And since we have a formula for the first "n" integers and not for some random range of integers we try to reduce what we're asked to find to this form

Hope this answers your question.

even integers between 99 and 301 are 100, 102, 104.....,300

for an arithmatic progression,a1,a2,a3....,an the sum is n* (a1+an)/2

so here.. n = 101, hence sum = 101 * 400/2 = 101 * 200 = 20200 : B

Sum = 100+102+...+300
= 100*101 + (0+2+4+...+200) [Taking 100 from each term in series, there are 101 terms]
= 100*101 + 2*(1+2+..+100) [Taking 2 common]
= 100*101 + 2*100*101/2 [Using formula given]
= 2*100*101
= 20200

Answer is (b)

can someone explain me why do we subtract 49 and no 50 ?

The sum of n consecutive odd integers starting from 1 is \(n^2\)

e.g. sum of 1 + 3 + 5 + 7 = \(4^2\)

GMAT doesn't expect you to know it but it is a standard formula and doesn't hurt to know.
It can be easily derived using n(n+1)/2 formula. Let's wait for a day or two to see if someone comes up with a simple method of doing it.


We need to find the sum
50 + 51 + 52 +...+ 150 (The sum we need includes 50)

It can be done in two ways:

1. Using n(n + 1)/2

1 + 2 + 3 + ...49 + 50 + 51 + ....150 = 150.151/2
Also, 1 + 2 + 3 + ....+ 49 = 49.50/2

To get the required sum, we just subtract second equation from first.
We get: 50 + 51 + ...150 = 150.151/2 - 49.50/2

2. Using average of AP concept.
Note 50, 51...150 is an arithmetic progression. (Common difference between adjacent terms.)
The average of this AP = (first term + last term)/2 = (50 + 150)/2 = 100
Sum of AP = Average * No of terms = 100 * 101

Ok. Let's quickly review sum of some series:

- Sum of positive consecutive integers starting from 1 is n(n + 1)/2 where n is the number of integers.
1+2+3+4+5+...+10 = 10*11/2

- Sum of positive consecutive even integers starting from 2 is n(n + 1) where n is the number of integers.
2+4+6+8+10 = 5*6 = 30

The explanation is simple.
2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common
Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(n^2\)
1+3+5+7+9+11 = \(6^2\)

I can derive it in the following way:
1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - n
I add and subtract n from the right side.
The right side becomes: 2+4+6+8+10+12 - n = n(n+1) - n = \(n^2\)

So 100 + 102 + 104 + 106 + ...+ 300 can be solved in multiple ways.

It has been solved above in the following manner:

100 + 102 + 104 + 106 + ...+ 300 = 2( 50 + 51 + 52 + 53 + ...+ 150) Take out 2 common and find the sum in brackets.

50 + 51 + 52 + 53 + ...+ 150
We know the sum of consecutive integers but only when they start from 1. So what do we do?

We find the sum of first 150 numbers and subtract the sum of first 49 numbers from it. That will give us the sum of numbers from 50 to 150. Note that we subtract 49 numbers because 50 is part of our series. We need it so we do not subtract it.

50 + 51 + 52 + 53 + ...+ 150 = 150*151/2 - 49*50/2

Then 100 + 102 + 104 + 106 + ...+ 300 = 2( 150*151/2 - 49*50/2) = 20200

There is an easier way (without using formula) :

number of even integers from 99 and 301 = (300-100)/2 + 1 = 101
Average = (100+300)/2 = 200

Sum of all even integers between 99 and 301 = 101 * 200 = 20,200

If there are 100 even integers in the sequence why do you +1 and multiply the average (200) by 101?

Question: even integers between 99 and 301

We know 99 is not even and 301 is not even; so ignore both of those

The actual sequence becomes:

100,102,104,106,108,.......,296,298,300

In the above sequence; the first number 100 is divisible by 2(or is even) and the last number 300 is also divisible by 2(or is even).

The common difference "d" is 2.

For such sequences; there is a formula to find the number of elements:

It is {(Last Number - First Number)/d} + 1

So; the above sequence actually contains {(300-200)/2}+1 = 101 elements

And Average = (first number + last number)/2 = (100+300)/2 = 200

Sum = Average * Number of Elements = 101 * 200 = 20200

Ans: 20200