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For every integer k from 1 to 10, inclusive, the kth term of

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For every integer k from 1 to 10, inclusive, the kth term of [#permalink]

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New post 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


[Reveal] Spoiler:
Shall I use the formula for Geometric Progression;

Sn = a*[ (1-r^n)/(1-r) ]

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: for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Aug 2013, 06:24, edited 2 times in total.
Added OA.
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 [#permalink]

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New post 03 Jun 2007, 08:40
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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ANS [#permalink]

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New post 03 Jun 2007, 12:20
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=1/2-1/4+1/8-1/16+...........
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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Re: ANS [#permalink]

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New post 03 Jun 2007, 16:29
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=1/2-1/4+1/8-1/16+...........
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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New post 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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New post 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 \(= \frac{1}{2}-\frac{1}{2^2}.......-\frac{1}{2^{10}}\)

\(=[ \frac{1}{2}+\frac{1}{2^3}]+... -[\frac{1}{2^2}+\frac{1}{2^4}...]\)

\(=[ \frac{1}{2}+\frac{1}{2^3}]+... -\frac{1}{2}[\frac{1}{2}+\frac{1}{2^3}...]\)

\(=\frac{1}{2}[ \frac{1}{2}+\frac{1}{2^3}+. . . . . . +\frac{1}{2^9}...]\) =\(\frac{1}{2}[0.5+0.125+0.0xxx+..] \approx =0.3125.\)

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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]

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New post 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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New post 26 Feb 2016, 20:09
it can be solved by simply by G.P formula
\(sum=\frac{a(1-r^n)}{1-r}\)
here \(r= \frac{-1}{2}\)
a= \(\frac{1}{2}\)

sum=\(\frac{1}{2[([fraction]1-(-[fraction]1/2})^10)/1-(1/2)[/fraction]]\)
= \(\frac{1}{2}[2*(\frac{1024-1)}{3*1024}]\)
=\(\frac{1023}{3*1024}\)
approx .3

I hope i am right
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For every integer k from 1 to 10, inclusive, the kth term of   [#permalink] 26 Feb 2016, 20:09
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