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A sequence of numbers (geometric sequence) is given by the [#permalink]
19 Jan 2013, 10:53

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Question Stats:

35% (07:06) correct
65% (02:30) wrong based on 316 sessions

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]
19 Jan 2013, 11:17

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maglian wrote:

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

So, we need to find the least value of n for which \frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000} --> 2^n>7500 --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]
19 Jan 2013, 11:23

Bunuel wrote:

So, we need to find the least value of n for which \frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000} --> 2^n>7500 --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]
11 Oct 2013, 12:06

Though I got it wrong due to wrong approximation, I found an alternate method.

Take n=1, so G(1) - G(2) = 5(-1/2)^1 - 5(-1/2)^2. On solving we get some number upon power of 2 as 2( since 2^2 is LCM)

Since the value compared is with 1000, we need to take values of 2 above index 10

Take option C: | G(12) - G(13) | = 5(-1/2)^12 - 5(-1/2)^13 , solve it to get 5/2^13. If we take split denominator as 2^3 and 2^10 we get (5/2^3)*(1/2^10).

5/8 = 0.625 and 1/2^10 = 1024 , if you compare it with RHS 1/1000 we are almost there. I had approximated 1024 as 1000, and wrongly took 5/8 as 0.125 which gave me it less than 1000, but in actuals its more than 1/1000 So next number could be the answer, check for G13 and G14, you have the answer. Hope I am able to express it clearly

Honestly speaking I dont think such questions can come up in GMAT

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]
02 Jul 2014, 19:01

Bunuel wrote:

maglian wrote:

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

So, we need to find the least value of n for which \frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000} --> 2^n>7500 --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).