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Re: Brute Force or Some Pattern [#permalink]
01 Apr 2012, 22:18

1

This post received KUDOS

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). _________________

some one please explain how common difference is - 4

please include the complete series so that we can see how this works

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.

some one please explain how common difference is - 4

please include the complete series so that we can see how this works

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 _________________

Re: What is 10 - 8 + 6 - 4 + ... - (-20) ? [#permalink]
01 Nov 2014, 13:36

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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