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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 :-)

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