# Lesson: Sums of Sequences

## Comment on Sums of Sequences

### Hi Brent!

Hi Brent!

http://gre.myprepclub.com/forum/the-sum-of-the-odd-even-integers-1867.html

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/the-sum-of-the-odd-even-integers-1867.html

Hi Brent - besides the way that you solved it on the forum, can u provide an alternative way to solve it? Mainly because I'm having a hard time understanding the other solutions that were posted.

### You bet!

You bet!

I've added a second/different solution here: https://gre.myprepclub.com/forum/the-sum-of-the-odd-even-integers-1867.h...

Cheers,
Brent

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/the-sum-of-the-odd-even-integers-1867.html#p19621

I solved this question by using the formula Sum = n/2(first + last)

For A: n = (199-1+1) = 199/2 = 99.5 ~ 100
Sum = 100/2(1+199) = 10,000

For B: n = (198-2+1) = 197/2 = 98.5 ~ 99
Sum = 99/2(2+198) = 9,900

Is this approach correct?

That's a perfectly valid solution. Nice work!

Cheers,
Brent

### what lesson does this

what lesson does this equation fall under? Because I am still having a hard time understanding how you solved it?

### The concept required to

The concept required to answer this question falls under the heading Sums of Sequences (video above), but my solution somewhat circumvents the requirement to find any actual sums.

Also note that this is a very tricky question!

Cheers,
Brent

### I noticed that you have not

I noticed that you have not mentioned geometric sequences in this video. They aren't that important for gre or they out of scope?

### Good question.

Good question.

Geometric sequences are out of scope. That is, you aren't expected to know what a geometric sequence is, and you aren't expected to know how to find individual terms (or the sum) of a geometric sequence.

That said, it's conceivable that a GRE question might describe/define a sequence that is geometric. However, the test-makers would likely construct the question so that previous knowledge of geometric sequences would not give the test-taker an advantage.

Cheers,
Brent

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/2-5112.html

In the solutions for the question posted above, looking at the explanations I understand everything until step 2 (or when you factor out a (-2) and leave the positive difference of squares)). Can you explain to me why you factored out a -2, and the steps after?

Thank you!

### Sorry. It looks like I made

Sorry. It looks like I made an error after factoring out the -2.

I've edited my response (at https://gre.myprepclub.com/forum/2-5112.html#p9241), but I'll explain my rationale nonetheless.

Let's start here: answer = (-2)(2 + 4) + (-2)(6 + 8) + (-2)(10 + 12) + (-2)(14 + 16) + (-2)(18 + 20)

Rather than multiply -2 by each of the sums and then find that entire sum, I see that I can save some time by first factoring out the -2 to get the following:
answer = (-2)[(2 + 4) + (6 + 8) + (10 + 12) + (14 + 16) + (18 + 20)]

From here, I just need to add all of the values inside the square brackets and then multiply the entire sum by -2

Does that help?

Cheers,
Brent

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/if-9189.html
can you please elaborate? I didnt understand how you got 21+22?

The given pattern, (-20) + (-19) + (-18) + (-17) + . . . , tells us that the terms are CONSECUTIVE integers.

In my solution, I recognized that, if we keep the sequence going to 20, we get: (-20) + (-19) + (-18) + (-17) + . . . + 19 + 20, which equals ZERO.

So, from that point, if we keep adding more CONSECUTIVE integers, we add 21, then we add 22, etc

We know that (-20) + (-19) + (-18) + (-17) + . . . + 19 + 20 = 0,
So, (-20) + (-19) + (-18) + (-17) + . . . + 19 + 20 + 21 = 0 + 21 = 21
And (-20) + (-19) + (-18) + (-17) + . . . + 19 + 20 + 21 + 22 = 0 + 21 + 22 = 43
And (-20) + (-19) + (-18) + (-17) + . . . + 19 + 20 + 21 + 22 + 23 = 66
And (-20) + (-19) + (-18) + (-17) + . . . + 19 + 20 + 21 + 22 + 23 + 24 = 90

Does that help?

Cheers,
Brent

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/the-sum-of-the-odd-even-integers-1867.html

I was wondering if this is a valid approach:

A = 199 numbers/2 = 99.5 = around 100 numbers are even
B = 197 numbers/2 = 98.5 = around 99 numbers are odd

Then use --> sum = n(n+1)/2

So:
A = 100(100+1)/2 = 5050
B = 99(99+1)/2 = 4950

So, A> B

This approach can get you into trouble at times.
For example, why did you use 197 for Quantity B?

Also, the sum = n(n+1)/2 applies only to consecutive integers.
It doesn't apply to consecutive EVEN integers or consecutive ODD integers.

See what answer you get if you consider the integers from 1 to 198.
You should find that A < B.
Give it a try.

Cheers,
Brent

### Hello !

Hello !

https://gre.myprepclub.com/forum/the-sum-of-the-odd-even-integers-1867.html
Please, can you explain a little more why you subtract both of them. And why you get positives ones
When I subtract it . I get many minus ones.

Also if it possible can you explain the first approach that is the beginning, in the same link.
Thank you

In my 2nd step, I subtracted all of Quantity B from both quantities so that I could have 0 in Quantity B (makes it very easy to compare)

Notice that, in the sum 1 + 3 + 5 + ... + 197 + 199, we are adding 100 numbers
Conversely, in the sum 2 + 4 + 6 + ... + 196 + 198, we are adding 99 numbers

So, if we subtract (as I believe you did), we get:
(1-2) + (3-4) + (5-6) + . . . .+ (195-196) + (197-198) + 199
= (-1) + (-1) + (-1) + . . . + (-1) + (-1) + 199
= -99 + 199
= 100

Does that help?

Cheers,
Brent

### Hi Brent, could you explain

Hi Brent, could you explain how we know that for the first we're adding 100 numbers and for the second 99? Thanks!

### Good question.

Good question.
There are 199 integers from 1 to 199 inclusive.
Since we start with an odd integer (1) and end with an odd integer (199), we can conclude that the number of odd integers must be one greater than the number of even integers.
So, there are 100 odd integers, add 99 even integers.

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/for-each-integer-n-1-let-a-n-denote-the-sum-2131.html
In this question why is it not solved by using the formula directly [n(n+1)]/2.

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/for-each-integer-n-1-let-a-n-denote-the...

Good question!
My solution assumed that students are unaware of that formula. I think it's useful to see how we can still answer the question without any knowledge of the formula.

That said, I just added a second solution by applying the formula: https://gre.myprepclub.com/forum/for-each-integer-n-1-let-a-n-denote-the...

Cheers,
Brent

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/if-9189.html
I don't get why you would equal 24 to -2x-4. Can you explain?
Thanks

### My solution: https:/

Let's start here: 21 + 22 + ? + ? + ... + (-2x - 4) = 90
Starting from 21, we want to keep adding consecutive integers until we get a sum of 90.
NOTE: (-2x - 4) is the last consecutive integer in the sum

After testing a few sums, we find that 21 + 22 + 23 + 24 = 90
So the last consecutive integer in the sum is 24
This means -2x - 4 = 24

Does that help?

Hi Brent,