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For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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03 Jun 2007, 03:27
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For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1)*(1/2^k). If T is the sum of the first 10 terms in the sequence, then T is A. greater than 2 B. between 1 and 2 C. between 0.5 and 1 D. between 0.25 and 0.5 E. less than 0.25 Shall I use the formula for Geometric Progression;
Sn = a*[ (1r^n)/(1r) ]
and calculate finally up to Sn = ( 2^10  1 )/ (3* (2^10)) = 1023/3*1024
This seems to be such a long and tedious calculation!! and I don't think we need to know formulae for Geometric Progressions etc.
What should be fast approach to solve this problem? Please explain. OPEN DISCUSSION OF THIS QUESTION IS HERE: foreveryintegerkfrom1to10inclusivethekthtermof88874.html
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Last edited by Bunuel on 11 Aug 2013, 06:24, edited 2 times in total.
Added OA.



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LM,
i am afraid but this is what u are asked to do. right method
But u don't need to calculate A= 1023/3*1024. Just A is almost same with
1024/3*1024 = 1/3 (ROUND UP unit digit)
Vote for D



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suppose
S=1+1/2+1/4+1/8............................ this is an infinite GP and here
s comes to be S=2(which is max).
Now we are given
S=1/21/4+1/81/16+...........
here solving for first two terms we get 1/4 i.e .25 means our answer should be greater than .25 as all other terms gives positive in pair of two.
Now the max. value of this expansion will be 1 when all the terms are positive and upto infinite but in this every alternate term is negative thus reducing the sum of expansion to half of 1 that is 1/2(.5) which can be the max. value.
Thus value lies b/w .25 and .50.
D should be the answer.



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deepakguptaeng@gmail.com wrote: suppose S=1+1/2+1/4+1/8............................ this is an infinite GP and here s comes to be S=2(which is max). Now we are given S=1/21/4+1/81/16+........... here solving for first two terms we get 1/4 i.e .25 means our answer should be greater than .25 as all other terms gives positive in pair of two. Now the max. value of this expansion will be 1 when all the terms are positive and upto infinite but in this every alternate term is negative thus reducing the sum of expansion to half of 1 that is 1/2(.5) which can be the max. value. Thus value lies b/w .25 and .50.
D should be the answer.
Good explanation. Answer is correct. But please remember it is not "infnite G.P." Up to 10 terms only.



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For every integer k from 1 to 10, inclusive, the kth term... [#permalink]
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04 Aug 2013, 12:05
For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1) (1/2^k). If T is the sum of the first 10 terms in the sequence, then T is
a) greater than 2 b) between 1 and 2 c) between 1/2 and 1 d) between 1/4 and 1/2 e) less than 1/4
Last edited by Zarrolou on 04 Aug 2013, 12:07, edited 1 time in total.
Merging similar topics.



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Re: For every integer k from 1 to 10, inclusive, the kth term... [#permalink]
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04 Aug 2013, 12:24
njkhokh wrote: For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1) (1/2^k). If T is the sum of the first 10 terms in the sequence, then T is
a) greater than 2 b) between 1 and 2 c) between 1/2 and 1 d) between 1/4 and 1/2 e) less than 1/4 From the given sum, T \(= \frac{1}{2}\frac{1}{2^2}.......\frac{1}{2^{10}}\) \(=[ \frac{1}{2}+\frac{1}{2^3}]+... [\frac{1}{2^2}+\frac{1}{2^4}...]\) \(=[ \frac{1}{2}+\frac{1}{2^3}]+... \frac{1}{2}[\frac{1}{2}+\frac{1}{2^3}...]\) \(=\frac{1}{2}[ \frac{1}{2}+\frac{1}{2^3}+. . . . . . +\frac{1}{2^9}...]\) =\(\frac{1}{2}[0.5+0.125+0.0xxx+..] \approx =0.3125.\) D.
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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11 Aug 2013, 06:24



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For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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26 Feb 2016, 20:09
it can be solved by simply by G.P formula \(sum=\frac{a(1r^n)}{1r}\) here \(r= \frac{1}{2}\) a= \(\frac{1}{2}\) sum=\(\frac{1}{2[([fraction]1([fraction]1/2})^10)/1(1/2)[/fraction]]\) = \(\frac{1}{2}[2*(\frac{10241)}{3*1024}]\) =\(\frac{1023}{3*1024}\) approx .3 I hope i am right
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