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]

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*[ (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.

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[#permalink]

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]

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]

03 Jun 2007, 16:29

deepakguptaeng@gmail.com wrote:

Good explanation. Answer is correct. But please remember it is not "infnite G.P." Up to 10 terms only.

Re: ANS [#permalink] 03 Jun 2007, 16:29

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

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

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