# If the sum of the consecutive integers from -22 to x, inclusive, is 72

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• Apr 26th 2008, 04:59 PM
Potterpie
If the sum of the consecutive integers from -22 to x, inclusive, is 72
then what is the value of x?

this question is from my SAT study guide. Its difficulty is hard.

Thanks so much!

• Apr 26th 2008, 05:05 PM
icemanfan
Quote:

Originally Posted by Potterpie
then what is the value of x?

this question is from my SAT study guide. Its difficulty is hard.

Thanks so much!

For this problem, first consider: what is the sum of consecutive integers from -22 to 22?
• Apr 26th 2008, 05:06 PM
TKHunny
-22 + -21 + -20 + ... + x = 72

Obviously, x > 0, otherwise you'll never get to 72 after starting negative.

You should have some familiarity with the "Arithmetic Sequence"

-22 = -22

-21 = -22 + 1

-20 = -22 + 1(2)

-19 = -22 + 1(3)

...

x = -22 + 1(22+x)

Where does that leave us?
• Apr 26th 2008, 05:06 PM
Potterpie
Wouldn't it be 44?

@Hunny: I haven't learned arithmetic sequence yet because I haven't finished algebra II. unless I was supposed to learn it by now.

But your last equation, x=-22+1(22+x), simplified, just leaves me with x=x.
• Apr 26th 2008, 05:11 PM
icemanfan
Quote:

Originally Posted by Potterpie
Wouldn't it be 44?

I was thinking of breaking down the sum like this:

$\displaystyle -22 + (-21) + (-20) + ... + 0 + 1 + 2 + ... + 22$ is the same as

$\displaystyle -22 + 22 + (-21) + 21 + (-20) + 20 + ...$
• Apr 26th 2008, 05:11 PM
Gusbob
Edit: see the post above, if you haven't learnt arithmetic series the above way is the best.
• Apr 26th 2008, 05:13 PM
icemanfan
Quote:

Originally Posted by Gusbob
$\displaystyle S_n = \frac{n}{2} (2a + (n-1)(d))$ where n = number of terms, a = 1st term (-22), and d(1) is the common difference

$\displaystyle \frac{n}{2} [2(-22) + (n-1)(1) ]=72$

Simplifying and Expanding the brackets

$\displaystyle \frac{-45n}{2} + \frac{n^2}{2} = 72$

$\displaystyle n^2 -45n -144 = 0$

$\displaystyle n = \frac{45\pm 51}{2}$

n cannot be negative, therefore n = 48

There are 48 terms in this series.

-22 + 48 = 26

x = 26

Check your work on that, Gusbob. The series of consecutive integers from -22 to 26 actually includes 49 terms.
• Apr 26th 2008, 05:16 PM
Gusbob
Quote:

Originally Posted by icemanfan
Check your work on that, Gusbob. The series of consecutive integers from -22 to 26 actually includes 49 terms.

Yeah, I noticed that and I noticed your post, which is a much simpler way of doing this.
• Apr 26th 2008, 05:17 PM
Potterpie
Okay, iceman's post made sense but I'm pretty sure we're not supposed to go through and add every single one. the SAT is timed.

Gusbob, that equation works once you change 48 to 49 (because I know the answer is 25, I just don't know how to get it), but where did you come up with that?

and hunny's equation just left me with x=x. You distribute the 1 into the parenthesis and then you end up with x=-22+22+x.
• Apr 26th 2008, 05:19 PM
Gusbob
Iceman's post is suggesting that the sum of -22 to 22 is 0

(-22 + 22) + (-21 + 21)+ ...+ (-1 + 1) + 0 = 0
• Apr 26th 2008, 05:20 PM
icemanfan
Quote:

Originally Posted by Potterpie
Okay, iceman's post made sense but I'm pretty sure we're not supposed to go through and add every single one. the SAT is timed.

Gusbob, that equation works once you change 48 to 49 (because I know the answer is 25, I just don't know how to get it), but where did you come up with that?

and hunny's equation just left me with x=x. You distribute the 1 into the parenthesis and then you end up with x=-22+22+x.

My point is that you get a bunch of terms that cancel out when you add the numbers from -22 to 22. It's the numbers you add in after that that make the sum of 72. It's as simple as $\displaystyle 23 + 24 + 25 = 72$
• Apr 26th 2008, 05:23 PM
Potterpie
I just calculated it. If you add up -22+-21, all the way to -1, the sum is not 0. It's -253 because you add consecutive negative integers.

I just have to figure out which number, x, to go up to to make the sum 72 instead of -253. <-- or wait, nvm. I don't know. Is that right? All I know is x=25. D:
• Apr 26th 2008, 05:27 PM
icemanfan
Quote:

Originally Posted by Potterpie
I just calculated it. If you add up -22+-21, all the way to -1, the sum is not 0. It's -253 because you add consecutive negative integers.

I just have to figure out which number, x, to go up to to make the sum 72 instead of -253. <-- or wait, nvm. I don't know. Is that right? All I know is x=25. D:

However, the sum from -22 to 22 is 0. You're making the problem harder than it needs to be.
• Apr 26th 2008, 05:29 PM
Potterpie
Okay, true. Right.

Then where do I go from that to get x=25?
• Apr 26th 2008, 05:42 PM
icemanfan
Quote:

Originally Posted by Potterpie
Okay, true. Right.

Then where do I go from that to get x=25?

The total sum has to be 72. When you add 23, you get a sum of 23. So how many more numbers do you have to add to get a sum of 72?
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