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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)
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)
maglian
\(|g(n) - g(n+1)| =|5 *(-\frac{1}{2})^n-5 *(-\frac{1}{2})^{n+1}|\) --> factor out \(5 *(-\frac{1}{2})^n\):
\(|5 *(-\frac{1}{2})^n-5 *(-\frac{1}{2})^{n+1}|=|5 *(-\frac{1}{2})^n*(1-(-\frac{1}{2})^1)|=|5 *(-\frac{1}{2})^n*\frac{3}{2}|=\frac{15}{2}*(\frac{1}{2})^n\).
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).
Answer: D.
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General Discussion
Bunuel
thank you. very simple explanation. you rock.
Great Explantion! Thanks bunuel
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
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
Posts: 3
We can write g(n+1) = -1/2*g(n)
so g(n) - g(n+1) = g(n) +1/2*g(n) = 3/2 g(n)
3/2 g(n) < 1/1000
or g(n)<1/1500
(1/2)^n<1/7500 (ignoring the -ve sign due to modulus)
n=13 is the least value which satisfies the equation.
Thanks,
Chirag Bhagat
so g(n) - g(n+1) = g(n) +1/2*g(n) = 3/2 g(n)
3/2 g(n) < 1/1000
or g(n)<1/1500
(1/2)^n<1/7500 (ignoring the -ve sign due to modulus)
n=13 is the least value which satisfies the equation.
Thanks,
Chirag Bhagat
Posts: 27
Bunuel
HOW DID YOU GET TO 2^(n) < 7500 exactly?
BestGMATEliza
This was great but where does 2 * 2^n come from?
Sorry
Bunuel
Hi Bunuel,
I got answer C, and also backed it up by calculation with a calculator.
it comes down to |-5*(1/2)^n +5*(1/2)*(n+1)| = |-5*(1/2)^(n+1)| < 1/1000
if (n+1) = 13, we get the answer we were looking for, so the answer should be C.
Are you sure that the answer needs to be D?
Can you find something wrong with my logic/calculations?
ronr34
You cannot factor out -1 the way you did. (-1/2)^n does not equal to -1*(1/2)^n if n is even, and we don't know whether it's even or odd.
The whole expression comes down to \(\frac{15}{2}*(\frac{1}{2})^n\).
Also, PLEASE read this: rules-for-posting-please-read-this-before-posting-133935.html#p1096628
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Posts: 80
maglian
the sequence is 15/4, 15/8, 15/16,......,15/2048, 15/4096, 15/ 8192, 15/16384 and so on
so the answer is (D)
Posts: 19
In the last step how did you come to the figure 2^13?
how do we arrive at 2 raised to the power of such a big number? esp the choice gives us close options
how do we arrive at 2 raised to the power of such a big number? esp the choice gives us close options
abrakadabra21
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Last visit: 10 Nov 2017
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Schools: Yale '18 ISB '18 (D) Judge"18
Schools: Yale '18 ISB '18 (D) Judge"18
Posts: 240
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)
5 *(-1/2)^n - 5 *(-1/2)^n+1
((-1/2)^n - (-1/2)^n+1 ) < 1/2^3*5^3 (1/5)
one of will be -ve and other positive so we are looking at the summation.
1/2^n (1+1/2)
2^n = 1024*5*3/2 (16 = around 2*2*2)
n =13
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)
5 *(-1/2)^n - 5 *(-1/2)^n+1
((-1/2)^n - (-1/2)^n+1 ) < 1/2^3*5^3 (1/5)
one of will be -ve and other positive so we are looking at the summation.
1/2^n (1+1/2)
2^n = 1024*5*3/2 (16 = around 2*2*2)
n =13
Why does |5∗(−1/2)^n− 5∗(−1/2)^n+1| becomes to/equals |5∗(−1/2)^n∗(1−(−1/2)^1)| ??
I don't understand why 5∗(−1/2)^n+1 can be expressed as n∗(1−(−1/2)^1
It would be nice if someone can explain.
I don't understand why 5∗(−1/2)^n+1 can be expressed as n∗(1−(−1/2)^1
It would be nice if someone can explain.
maglian
Let's see what the sequence is like:
g(n)=5 *(-1/2)^n
g(1) = -5/2
g(2) = 5/4
g(3) = -5/8
g(4) = 5/16
etc
|g(n) - g(n+1)| would be the absolute value of difference between two such consecutive terms. Consecutive terms will be of opposite signs so they will actually get added. Since we are taking their absolute value, if they have a negative sign, it will become positive.
So all we need to think about is that the sum of absolute values of two consecutive terms should be less than 1/1000
1/1000 = 5/5000
2^12 = 4096
So the power of 2 should be more than 12.
5/2^13 + 5/2^14 = 5/8192 + 5/16384 = 15/16384
We see that if we divide 16384 by 15, we will get something more than 1000.
Answer (D)
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Bunuel
can you please clarify how to get \(2^n>7500\)?
Im stuck with \(\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}\).
thank you!
omavsp
\(\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}\).
\(\frac{15*1000}{2}*(\frac{1}{2})^n<1\).
\(7500*\frac{1}{2^n}<1\).
\(7500<2^n\).
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I tried this way.
|g(n)-g(n+1)|=1/1000=1*10^-3=0.0001;
g(1)=-5/2
g(2)=5/8
absolute difference between two values=|-5/2-5/8|=-15/8
similarly, g(10)=5/1024
g(11)=-5/2048
difference=|5/1024- (-5/2048)|=5/1024+5/2048=15/2048;
The numerator is always 15 for all 'n'
15/1000 will yield 3 decimal point. But to be of lower value than 0.0001, we need four zeroes after decimal.
To get that, 15 must be divided by a value greater than 15000;
possible value is 2^14. so Option is D)
Please correct me if am wrong
|g(n)-g(n+1)|=1/1000=1*10^-3=0.0001;
g(1)=-5/2
g(2)=5/8
absolute difference between two values=|-5/2-5/8|=-15/8
similarly, g(10)=5/1024
g(11)=-5/2048
difference=|5/1024- (-5/2048)|=5/1024+5/2048=15/2048;
The numerator is always 15 for all 'n'
15/1000 will yield 3 decimal point. But to be of lower value than 0.0001, we need four zeroes after decimal.
To get that, 15 must be divided by a value greater than 15000;
possible value is 2^14. so Option is D)
Please correct me if am wrong
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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 ?
|g(n) - g(n+1)| = |5*(-1/2)^n - 5*(-1/2)^{n+1}| = 5*|(-1/2)^n (1-(-1/2)| = 5*3/2*|(-1/2)^n| = 15/2^{n+1}|(-1)^n| = 15/2^{n+1} < 1/1000
2^{n+1} > 15000
2^14 = 16384> 15000
n+1 = 14
n = 13
g(13),g(14)
IMO D
|g(n) - g(n+1)| = |5*(-1/2)^n - 5*(-1/2)^{n+1}| = 5*|(-1/2)^n (1-(-1/2)| = 5*3/2*|(-1/2)^n| = 15/2^{n+1}|(-1)^n| = 15/2^{n+1} < 1/1000
2^{n+1} > 15000
2^14 = 16384> 15000
n+1 = 14
n = 13
g(13),g(14)
IMO D
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isn't it a modular operation
so won't
-1/1000 < g(n) - g(n+1) < 1/1000 to be considered??
so won't
-1/1000 < g(n) - g(n+1) < 1/1000 to be considered??
Bunuel
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