What is 10 - 8 + 6 - 4 + ... - (-20) ?

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What is 10 - 8 + 6 - 4 + ... - (-20) ? [#permalink]

01 Apr 2012, 22:21

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Question Stats:

74% (02:29) correct

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based on 67 sessions

What is 10 - 8 + 6 - 4 + ... - (-20) ?

A. 8
B. 10
C. 12
D. 14
E. 16

[Reveal] Spoiler: OA

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Re: Brute Force or Some Pattern [#permalink]

01 Apr 2012, 23:12

GMATPASSION wrote:

What is 10 - 8 + 6 - 4 + ... - (-20) ?

8
10
12
14
16

I solved this by brute force? Any other way.

Pair them up.

10 - 8 = 2
6 - 4 = 2
2 - 0 = 2
.
.
-18 - (-20) = 2

There are a total of 16 terms (2*5 to 2*-10 gives you 5-(-10)+1 = 16 terms) so you will get 8 pairs. Hence the sum will be 2*8 = 16
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Re: Brute Force or Some Pattern [#permalink]

01 Apr 2012, 23:18

Note that 10-8 =2 and 6-4 =2. The next two terms, i.e. 2 and 0 also yield 2. Therefore the series is simply the sum of 2 over many terms.

How many such terms are there? The first negative term is 8 and the last is -20, with a common difference of -4. Alternatively, we could calculate the number of terms by taking the first term as 10, the common difference as -4, and the last term as -18.

-20 = 8 + (n-1) (-4)
=> n = 8

Therefore the answer is simply 2*8 = 16 , or option (E).
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Re: Brute Force or Some Pattern [#permalink]

01 Apr 2012, 23:20

GMATPASSION wrote:

What is 10 - 8 + 6 - 4 + ... - (-20) ?

8
10
12
14
16

I solved this by brute force? Any other way.

Or consider them as sum of 2 arithmetic progressions (though it would be much better if you see the pairing pattern)

I) 10 + 6 + 2 + (-2) ....+(-18)

AP with first term = 10 and common diff = -4
Sum = (8/2)(2*10 + 7*(-4)) = -32

II) -8 -4 - 0 - (-4) +..- (-20)
AP with first term = -8 and common diff = 4
Sum = (8/2)(2*(-8) + 7*(4)) = 48

Sum of the series = -32 + 48 = 16
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Re: Brute Force or Some Pattern [#permalink]

01 Apr 2012, 23:34

Yep - I was gonna post the same reply below. there are 2 APs here, you can just sum them
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Re: What is 10 - 8 + 6 - 4 + ... - (-20) ? [#permalink]

09 Jun 2012, 12:27

some one please enlighten me

please list the whole series and how a common difference of -4 is obtained .

Sorry my limited wisdom is unable to grasp the process

as far as I can see the terms are decreasing by 2 with alternating + and - sign

so the series should continue as 10 -8 +6 - 4 + 2 - 0 + (-2) - ( -4 ) + ( -6) - (-8 ) + ( -10) -( -12) +( -14 )

- (-16) +(-18)- (-20)

10 - 8 + 6 - 4 + 2 - 0 -2 + 4 - 6 + 8 - 10 + 12 - 14 + 16 - 18 + 20

some one please explain how common difference is - 4

please include the complete series so that we can see how this works
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Re: What is 10 - 8 + 6 - 4 + ... - (-20) ? [#permalink]

09 Jun 2012, 12:56

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stne wrote:

Here you go:

The sequence is 10-8+6-4+2-0+(-2)-(-4)+(-6)-(-8)+(-10)-(-12)+(-14)-(-16)+(-18)-(-20).

Notice that the odd numbered terms (1st, 3rd, 5th...) form arithmetic progression with common difference of -4 and the even numbered terms (2nd, 4th...) form arithmetic progression with common difference of 4:

The sum of the odd numbered terms is 10+6+2+(-2)+(-6)+(-10)+(-14)+(-18)=10+6+2-2-6-10-14-18=-32;

The sum of the even numbered terms is -8-4-0-(-4)-(-8)-(-12)-(-16)-(-20)=-8-4-0+4+8+12+16+20=48;

Their sum is -32+48=16.

Though I wouldn't recommend to solve this question this way. It's better if you notice that we have 8 pairs:
10-8=2;
6-4=2;
2-0=2;
(-2)-(-4)=2;
(-6)-(-8)=2;
(-10)-(-12)=2;
(-14)-(-16)=2;
(-18)-(-20)=2;

So, the sum of each pair is 2, which makes the whole sum equal to 8*2=16.

Hope it's clear.
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Re: What is 10 - 8 + 6 - 4 + ... - (-20) ? [#permalink]

09 Jun 2012, 13:30

Bunuel wrote:

stne wrote:

Here you go:

The sequence is 10-8+6-4+2-0+(-2)-(-4)+(-6)-(-8)+(-10)-(-12)+(-14)-(-16)+(-18)-(-20).

Notice that the odd numbered terms (1st, 3rd, 5th...) form arithmetic progression with common difference of -4 and the even numbered terms (2nd, 4th...) form arithmetic progression with common difference of 4:

The sum of the odd numbered terms is 10+6+2+(-2)+(-6)+(-10)+(-14)+(-18)=10+6+2-2-6-10-14-18=16;

The sum of the even numbered terms is -8-4-0-(-4)-(-8)-(-12)-(-16)-(-20)=-8-4-0+4+8+12+16+20=48;

Their sum is -32+48=16.

Though I wouldn't recommend to solve this question this way. It's better if you notice that we have 8 pairs:
10-8=2;
6-4=2;
2-0=2;
(-2)-(-4)=2;
(-6)-(-8)=2;
(-10)-(-12)=2;
(-14)-(-16)=2;
(-18)-(-20)=2;

So, the sum of each pair is 2, which makes the whole sum equal to 8*2=16.

Hope it's clear.

Thank you , now the previous explanations make sense

Just a small typo in your answer , I think it should be - 32 instead of 16 ( addition of odd terms ) , please do edit , to prevent confusion.

I was not able to see that even and odd terms are making an AP .

as Kudos is a better way of saying thank you , so you have been awarded , will be needing more of your assistance
in the future
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Re: What is 10 - 8 + 6 - 4 + ... - (-20) ? [#permalink]

09 Jun 2012, 13:38

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stne wrote:

just rearrange the terms you have written with putting all the terms subtracted together and the ones added together.

10 -(8) +6 -( 4) + 2 - (0) + (-2) - ( -4 ) + ( -6) - (-8 ) + ( -10) -( -12) +( -14 ) - (-16) +(-18)- (-20)

=> 10 +6 + 2 +(-2) +( -6) +(-10)+( -14 ) +(-18) -(8) -( 4) - (0) -( -4 )- (-8 )-( -12) - (-16)- (-20)

Hope that helps.

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Re: What is 10 - 8 + 6 - 4 + ... - (-20) ? [#permalink]

17 Jun 2012, 12:52

The quickest strategy for me was to pair them up and add the 2s.

10-8=2;
6-4=2;
2-0=2;
(-2)-(-4)=2;
(-6)-(-8)=2;
(-10)-(-12)=2;
(-14)-(-16)=2;
(-18)-(-20)=2;

2+2+2+2+2+2+2+2= 2*8 = 16

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Re: What is 10 - 8 + 6 - 4 + ... - (-20) ? [#permalink]

21 Dec 2012, 07:48

GMATPASSION wrote:

What is 10 - 8 + 6 - 4 + ... - (-20) ?

A. 8
B. 10
C. 12
D. 14
E. 16

I wrote them down and started cancelling pairs (+) against (-) then I was left with 16.

Answer: E

Re: What is 10 - 8 + 6 - 4 + ... - (-20) ? [#permalink] 21 Dec 2012, 07:48

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