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(1) \(x<-1\): \(x+x^2>0\) (x<-1 meas that x^2>|x|), \(x^3+x^4>0\), ..., \(x^9+x^{10}>0\), so the sum is also more than zero. Sufficient.

(2) \(x^2>2\): even if x itself is negative then still as above: \(x+x^2>0\), \(x^3+x^4>0\), ..., \(x^9+x^{10}>0\), so the sum is also more than zero. Sufficient.

Hi! This is a geometric progression. Could you explain this using the sum formula?

Thanks!

Sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}-1)}{r-1}\), where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\).

So, in our case \(b=x\) and \(r=x\) --> \(1+(x+x^2+x^3+x^4+...+x^9+x^{10})=1+sum_{10}=1+\frac{x*(x^{10}-1)}{x-1}\).

(2) \(x^2>2\) --> \(x<-\sqrt{2}\) or \(x>\sqrt{2}\) --> so, either \(1+\frac{x*(x^{10}-1)}{x-1}=1+\frac{negative*positive}{negative}=1+positive=positive\) or \(1+\frac{x*(x^{10}-1)}{x-1}=1+\frac{positive*positive}{positive}=1+positive=positive\). Sufficient.

Thanks! I was actually considering a total of 11 terms, and b=1. Is that right?

You can do that. If you take 1 as the first term then the formula will be \(sum=\frac{b*(r^{n}-1)}{r-1}=\frac{1*(x^{11}-1)}{x-1}=\frac{x^{11}-1}{x-1}\).

(2) \(x^2>2\) --> \(x<-\sqrt{2}\approx{-1.4}\) or \(x>\sqrt{2}\approx{1.4}\) --> so, either \(\frac{x^{11}-1}{x-1}=\frac{negative}{negative}=positive\) or \(\frac{x^{11}-1}{x-1}=\frac{positive}{positive}=positive\). Sufficient.

Re: Is 1+x+x^2+x^3+....+x^10 positive? [#permalink]

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04 Jan 2011, 21:14

I went for the 11 second solution and picked A...

If I took time on it and actualyl solved it - it is clear that the answer is D.

I got tripped up be thinking to myself "x could have two values -> not sufficient"

solution would have been let x = 2 or -2. -2 or 2^10 = 1024. 11 terms in series. Avg value of terms is 1025 => sum is 1025 / 2 = positive -> Sufficient

Re: Is 1+x+x^2+x^3+....+x^10 positive? [#permalink]

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27 Apr 2014, 10:59

Bunuel wrote:

gmatpapa wrote:

Is \(1+x+x^2+x^3+....+x^{10}\)positive?

1. \(x<-1\) 2. \(x^2>2\)

Source: GMAT Club Hardest problems.

Is \(1+(x+x^2)+(x^3+x^4)+...+(x^9+x^{10})>0\)?

(1) \(x<-1\): \(x+x^2>0\) (x<-1 meas that x^2>|x|), \(x^3+x^4>0\), ..., \(x^9+x^{10}>0\), so the sum is also more than zero. Sufficient.

(2) \(x^2>2\): even if x itself is negative then still as above: \(x+x^2>0\), \(x^3+x^4>0\), ..., \(x^9+x^{10}>0\), so the sum is also more than zero. Sufficient.

Answer: D.

Hi Bunnel,

According to st1 (x<-1)

so can write this as

1+ (-2) + (-2)^2+ (-2)^3+ ---(-2)^10

can I use above as geometric progression. ( dont include 1 in seried we will add it lastly)

If I will use this in gp then I will get result as <0 bcz first term is -ve

I agree with @caioguima, this is not a very good problem. GMAT won't give you a problem where the statements are completely redundant, and the answer to the question is already a definite Yes. At least, I haven't seen an official GMAT question that follows this format, please correct me if anyone has seen such an example.

To expand on @caioguima, the expression 1+x+x^2+x^3+....+x^10 is positive for all values of x greater than or equal to zero. If we rewrite this expression using the geometric series format, it becomes (x^11-1)/(x-1)[skipping those details here], and if we now consider the case of x<0, then both numerator and denominator are negative, making the expression positive for all values of x<0. Therefore, 1+x+x^2+x^3+....+x^10 is positive for all values of x. And the statements become redundant at this stage, which I have never seen on the GMAT.

Re: Is 1+x+x^2+x^3+....+x^10 positive? [#permalink]

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15 May 2015, 20:23

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Re: Is 1+x+x^2+x^3+....+x^10 positive? [#permalink]

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18 Jun 2016, 12:30

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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