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## Comment on

Sums of Sequences## Can you please assist with

Yvonne

## Hi Yvonne,

Hi Yvonne,

Let's start with "I got the sum of the even integers (100-300) which is 201." If we ADD a bunch of even integers, the sum will be EVEN (201 is odd). Also, the SUM will be a lot bigger than 201 :-)

I believe you are saying that there are 201 integers from 100 to 300 inclusive. That part is correct. However, there are two problems with this information:

1) The question is asking for the EVEN integers from 100 to 300 inclusive. Your calculation (of 201) just tells us the integers (both odd AND even).

2) The formula n(n+1)/2 tells us the sum of the first n positive integers. For example, to find the sum 1+2+3+4+5+6, we see that n = 6 in this case. So, using the formula, the sum = (6)(6+1)/2 = 21. You are taking the number 201 (which represents something totally different), and you are trying to use the formula that finds the sum of the first n integers.

I provide 3 different solutions to that question here: http://www.beatthegmat.com/consecutive-integers-t275599.html

Cheers,

Brent

## Thanks Brent!

I see what I did wrong. I read all three of your solutions and it definitely makes more sense now.

Do you offer solutions to all of the OG 2016 questions on the beatthegmat.com website? Or just a select number of questions? I find the OG explanations to the problems make absolutely no sense at times.

Thanks!

-Yvonne

## I have answered quite a few

I have answered quite a few of the OG2016 questions on both Beat The GMAT and GMAT Club, but not all of them.

## Quote:

term(n) = 1/x - 1/(x+1)

What is the sum of the first 100 terms of this sequence?

A) 0

B) 1/101

C) 99/100

D) 100/101

E) 1

Sir can we solve the problem this problem by n(n+1)/2 formula

## No, that formula won't help

No, that formula won't help us here. Instead, we must recognize that almost all of our fractions cancel out. Here's what I mean....

1st term = 1/1 - 1/(1 + 1) = 1/1 - 1/2

2nd term = 1/2 - 1/(2 + 1) = 1/2 - 1/3

3rd term = 1/3 - 1/(3 + 1) = 1/3 - 1/4

4th term = 1/4 - 1/(4 + 1) = 1/4 - 1/5

.

.

.

99th term = 1/99 - 1/(99 + 1) = 1/99 - 1/100

100th term = 1/100 - 1/(100 + 1) = 1/100 - 1/101

So, the SUM = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + . . . + (1/99 - 1/100) + (1/100 - 1/101)

We can that the 2nd term in each bracket cancels out with the 1st term in the next bracket.

So, the SUM = 1/1 - 1/(100 + 1) = 1 - 1/101 = 100/101

Answer: D

## Brent, your story about Carl

Anyway, I tried using this approach to answer several question. It seems that Gauss' approach can be used to solve another problem such as calculate the SUM of consecutive multiple 5 starts from 4 and ends in 69. The tricky one maybe in calculate the number of integer there. I found that total number is 14, and the sum is 511.

It is true that we can use it to solve other variation of this problems?

Once again, thank you!

## You are correct; we can apply

You are correct; we can apply Gauss' concept to a variety of questions.

## Hi Brent,

we don't consider zero as positive number but in below question we consider it.

What is the sum of first 10 non-negative even integers?

https://gmatclub.com/forum/what-is-the-sum-of-first-10-non-negative-even-integers-a-40-b-60-c-242936.html

If question had asked us to calculate the sum of the first 10 even numbers, will we still consider zero?

## Question link: https:/

Question link: https://gmatclub.com/forum/what-is-the-sum-of-first-10-non-negative-even...

Be careful; this is a very common trap on the GMAT.

If I say that k is a NON-NEGATIVE integer. This does NOT mean that k must be a positive integer (many students will make this conclusion).

A non-negative number is a number that is not negative. Since 0 is not negative, we can say that 0 is a non-negative number.

ASIDE: 0 is neither positive nor negative.

------------------------------------------

Your 2nd question: "If the question had asked for the sum of first 10 even numbers, will we still consider zero?"

An official GMAT question would never ask us to determine the sum of the first 10 even numbers, because even numbers are without beginning and end.

EVEN integers: . . . -8, -6, -4, -2, 0, 2, 4, 6, 8, . . .

The ". . . " means the pattern continues forever.

That said, if the question asked "What is the sum of first 10 POSITIVE even numbers?", then we'd find the following sum: 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20

## Hi Brent,

Unfortunately, I don't even "get" the question (below) or where to start!

If the sum of the consecutive integers from -22 to x inclusive is 72, what is the value of x?

A. 23

B. 25.

C. 50

D.75

E. 94

## The question, rephrased, is

The question, rephrased, is asking:

If (-22) + (-21) + (-20) + (-19) + . . . + x = 72, what is the value of x?

To answer this question, we need to make some observations:

(-1) + 0 + 1 = 0

(-2) + (-1) + 0 + 1 + 2 = 0

(-3) + (-2) + (-1) + 0 + 1 + 2 + 3 = 0

(-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4 = 0

Notice that POSITIVE and NEGATIVE numbers that have the same MAGNITUDE give us a sum of 0.

Given this, we can see that:

(-22) + (-21) + (-20) + . . . + 20 + 21 + 22 = 0

So: (-22) + (-21) + (-20) + . . . + 20 + 21 + 22 + 23 = 23

And: (-22) + (-21) + (-20) + . . . + 20 + 21 + 22 + 23 + 24 = 47

And: (-22) + (-21) + (-20) + . . . + 20 + 21 + 22 + 23 + 24 + 25 = 72 (PERFECT)

So, x = 25

Answer: B

Cheers,

Brent

## Thanks Brent,

The original question stated: If the sum of the consecutive integers from -22 to x, inclusive 72.

Without the "equals" 72 it just sounded off. Anyways, thanks for clearing that up. However, with that information I still think I would of solved it as x + 23 = 72, which equals 49. I didn't get that it was looking for the next number in the sequence. I just assumed to solve for "x".

## Yeah, it's not a well-worded

Yeah, it's not a well-worded question.

In fact, I assumed you missed a word and edited your initial question :-)

## 2,345

2,354

2,435

+5,432

There are 24 different four-digit integers than can be formed using the digits 2, 3, 4 and 5 exactly once in each integer. The addition problem above shows 4 such integers. What is the sum of all 24 such integers?

24,444

28,000

84,844

90,024

93,324

## My full solution: https:/

My full solution: https://gmatclub.com/forum/there-are-24-different-four-digit-integers-th...

## Hi Brent, referring to the

For eg: -19,-18,-16........20,21,21?

## Great question!

Great question!

That formula applies only to the sum: 1 + 2 + 3 + . . . n

That said, we can sometimes find a way to rewrite a sequence so that we can apply the formula.

For example, let's say we want to find the sum of all positive multiples of 5 from 5 to 95 inclusive.

That is, we want to evaluate 5 + 10 + 15 + 20 + . . . + 90 + 95

Notice that 5 + 10 + 15 + 20 + . . . + 90 + 95 = 5(1 + 2 + 3 + . . . + 18 + 19)

= 5[(19)(19 + 1)/2]

= 950

Cheers,

Brent

## Great Explanation!Thanks!

## Hi Brent!

I am struggling with the following question: https://gmatclub.com/forum/what-is-the-sum-of-odd-integers-from-35-to-85-inclusive-221343.html

In your answer approach you wrote that the result must be divisible by 120.

I don't understand why this is true in the question, since the number of terms to be added is odd. So there has to be one number which does not match another and therefore does not add up to 120 right? So as I see it if a random number is added to the product of 120 and some number the sum is not necessarily divisible by 120.

Can you tell me where my mistake is? I hope you understand my question.

Thank you!

BR Pia

## https://gmatclub.com/forum

Question link: https://gmatclub.com/forum/what-is-the-sum-of-odd-integers-from-35-to-85...

Good question, Pia!

There are 26 odd integers from 35 to 85 inclusive.

This means there are 13 PAIRS of values that add to 120

So, the total sum = (13)(120) = 1560

ASIDE: Here's how I know there are 26 numbers in the sum:

Let's take the terms 35, 37, 39, . . . 81, 83, 85 and rewrite them as follows:

35 = 2(17) + 1

37 = 2(18) + 1

39 = 2(19) + 1

41 = 2(20) + 1

.

.

.

81 = 2(40) + 1

83 = 2(41) + 1

85 = 2(42) + 1

So, the number of ODD integers from 35 to 85 inclusive = the number of CONSECUTIVE integers from 17 to 42 inclusive

The number of CONSECUTIVE integers from 17 to 42 inclusive = 42 - 17 + 1 = 26

Does that help?

Cheers,

Brent

## Yes, thank you so much!

## Hi Brent,

I did a practice exam on the official gmatofficalprep.mba.com and got the following question first:

If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?

A: 10

B: 11

C: 12

D: 13

E: 14

I was unable to solve it, could you help me out?

Thanks in advance!

Cheers, Glenn

## Hi Glenn,

Hi Glenn,

Here's my full solution: https://gmatclub.com/forum/if-n-is-a-positive-number-gmat-prep-86490.htm...

Cheers,

Brent

## Hi Brent,

https://gmatclub.com/forum/in-a-sequence-each-term-is-obtained-by-adding-4-to-the-preceding-one-223579.html

I followed the same approach as shown by you except for the last part. After identifying the value of n, I took out the value of the 40th term = -10 + (39 times 4) = -10 + 156 = -146. Since now we know the 40th term, I then used the formula n(n+1)/2 but did not get the same answer. Can you please help me to understand what went wrong?

Thank you.

## If you show me your

If you show me your calculations, I can probably help.

Cheers,

Brent

## Sure!

term 40 = -10+(sum of 39 4's) = -10 + 156 = -146. Using the formula n(n+1)/2 = (-146)(-145)/2 = 21,170/2 = 10,585

but the correct answer is 2,720

Can you help me here, please?

## Question link: https:/

Question link: https://gmatclub.com/forum/in-a-sequence-each-term-is-obtained-by-adding...

There are a few issues with your solution.

First, the formula n(n+1)/2 tells us the sum of the CONSECUTIVE integers from 1 to n.

For example, 1+2+3+4+5 = (5)(6)/2 = 15

In the original question, the terms increase by 4 each time, which means they aren't consecutive (which means we can't apply the formula).

Also, the sequence starts at -10 (not 1), which means we can't directly apply the formula.

Finally, on a smaller note, term40 = -10 + 156 = 146 (not -146)

Does that help?

Cheers,

Brent

## Yes, it does! I'd thought we

Thanks so much, Brent!

## Hey Brent,

in this Q:

https://gmatclub.com/forum/what-is-the-sum-of-odd-integers-from-35-to-85-inclusive-221343.html

You presented a reeeally nice shortcut. However, if the number of integers had been odd, it would have not been possible, right?

Like taking 120 times some integer is only posible if there are pairs always?

Cheers,

Philipp

## Question link: https:/

Question link: https://gmatclub.com/forum/what-is-the-sum-of-odd-integers-from-35-to-85...

You're correct. If there were an odd number of terms, I wouldn't have been able to conclude the answer is divisible by 120.

Cheers,

Brent

## Hey Brent,

in this Q:

https://gmatclub.com/forum/what-is-the-sum-of-odd-integers-from-35-to-85-inclusive-221343.html

I understand that since the number of odd integers between 35 and 85 is an even number, you could club numbers together and multiplied 120 by this even number of terms. However, What would have been your approach is the number of integers had been odd?

## Question link: https:/

Question link: https://gmatclub.com/forum/what-is-the-sum-of-odd-integers-from-35-to-85...

If there's an odd number of values, the sum will be divisible by the middlemost number (aka the median).

That said, it's possible that more than 1 answer choice is divisible by the median, in which case, you'll need to apply a different strategy (like one of those used in the thread).