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# For every integer k from 1 to 10, inclusive the "kth term of a certain

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Joined: 16 Aug 2015
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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15 Jun 2017, 22:26
Bunuel wrote:
For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by $$(-1)^{(k+1)}*(\frac{1}{2^k})$$ 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

First of all we see that there is set of 10 numbers and every even term is negative.

Second it's not hard to get this numbers: $$\frac{1}{2}$$, $$-\frac{1}{4}$$, $$\frac{1}{8}$$, $$-\frac{1}{16}$$, $$\frac{1}{32}$$... enough for calculations, we see pattern now.

And now the main part: adding them up is quite a job, after calculations you'll get $$\frac{341}{1024}$$. You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than $$\frac{1}{4}$$ and less than $$\frac{1}{2}$$, so answer is D.

BUT there is shortcut:

Sequence $$\frac{1}{2}$$, $$-\frac{1}{4}$$, $$\frac{1}{8}$$, $$-\frac{1}{16}$$, $$\frac{1}{32}$$... represents geometric progression with first term $$\frac{1}{2}$$ and the common ratio of $$-\frac{1}{2}$$.

Now, the sum of infinite geometric progression with common ratio $$|r|<1$$, is $$sum=\frac{b}{1-r}$$, where $$b$$ is the first term.

So, if the sequence were infinite then the sum would be: $$\frac{\frac{1}{2}}{1-(-\frac{1}{2})}=\frac{1}{3}$$

This means that no matter how many number (terms) we have their sum will never be more then $$\frac{1}{3}$$ (A, B and C are out). Also this means that the sum of our sequence is very close to $$\frac{1}{3}$$ and for sure more than $$\frac{1}{4}$$ (E out). So the answer is D.

Other solutions at: http://gmatclub.com/forum/sequence-can- ... ml#p668661

Bunuel, how come the ratio here is -1/2, and not 1/2? If there are alternating signs in a sequence, is the "ratio" always the negative value? thanks
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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15 Jun 2017, 22:55
1
iyera211 wrote:
Bunuel wrote:
For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by $$(-1)^{(k+1)}*(\frac{1}{2^k})$$ 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

First of all we see that there is set of 10 numbers and every even term is negative.

Second it's not hard to get this numbers: $$\frac{1}{2}$$, $$-\frac{1}{4}$$, $$\frac{1}{8}$$, $$-\frac{1}{16}$$, $$\frac{1}{32}$$... enough for calculations, we see pattern now.

And now the main part: adding them up is quite a job, after calculations you'll get $$\frac{341}{1024}$$. You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than $$\frac{1}{4}$$ and less than $$\frac{1}{2}$$, so answer is D.

BUT there is shortcut:

Sequence $$\frac{1}{2}$$, $$-\frac{1}{4}$$, $$\frac{1}{8}$$, $$-\frac{1}{16}$$, $$\frac{1}{32}$$... represents geometric progression with first term $$\frac{1}{2}$$ and the common ratio of $$-\frac{1}{2}$$.

Now, the sum of infinite geometric progression with common ratio $$|r|<1$$, is $$sum=\frac{b}{1-r}$$, where $$b$$ is the first term.

So, if the sequence were infinite then the sum would be: $$\frac{\frac{1}{2}}{1-(-\frac{1}{2})}=\frac{1}{3}$$

This means that no matter how many number (terms) we have their sum will never be more then $$\frac{1}{3}$$ (A, B and C are out). Also this means that the sum of our sequence is very close to $$\frac{1}{3}$$ and for sure more than $$\frac{1}{4}$$ (E out). So the answer is D.

Other solutions at: http://gmatclub.com/forum/sequence-can- ... ml#p668661

Bunuel, how come the ratio here is -1/2, and not 1/2? If there are alternating signs in a sequence, is the "ratio" always the negative value? thanks

Common ratio is a ratio of consecutive terms. Divide two consecutive terms what do you get? 1/2 or -1/2?
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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16 Jun 2017, 07:05
1
Top Contributor
prathns 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

List some terms to see the pattern.

We get: T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
Notice that we can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)

Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less than 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024)
= 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, 1/4 < T < 1/2

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

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25 Jun 2017, 11:06
prathns 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

Refer to the solution in the picture
Attachments

Solution Kth Term.jpeg [ 24.64 KiB | Viewed 819 times ]

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

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26 Apr 2019, 00:54
Hi

Well we all understood that series is $$\frac{1}{2}$$ , $$\frac{-1}{2^2}$$ , $$\frac{1}{2^3}$$, $$\frac{-1}{2^4}$$, $$\frac{1}{2^5}$$ ,$$\frac{-1}{2^6}$$,$$\frac{1}{2^7}$$ ,$$\frac{-1}{2^8}$$ ,$$\frac{1}{2^9}$$ ,$$\frac{-1}{2^10}$$

Can i rewrite the series as

{$$\frac{1}{2}$$ +$$\frac{1}{2^3}$$ +$$\frac{1}{2^5}$$ +$$\frac{1}{2^7}$$ +$$\frac{1}{2^9}$$ } - { $$\frac{1}{2^2}$$ + $$\frac{1}{2^4}$$ + $$\frac{1}{2^6}$$+ $$\frac{1}{2^8}$$ + $$\frac{1}{2^10}$$ }

{$$\frac{1}{2}$$ +$$\frac{1}{2^3}$$ +$$\frac{1}{2^5}$$ +$$\frac{1}{2^7}$$ +$$\frac{1}{2^9}$$ } -$$\frac{1}{2}$$ { $$\frac{1}{2}$$ +$$\frac{1}{2^3}$$ +$$\frac{1}{2^5}$$ +$$\frac{1}{2^7}$$ +$$\frac{1}{2^9}$$ }

Let {$$\frac{1}{2}$$ +$$\frac{1}{2^3}$$ +$$\frac{1}{2^5}$$ +$$\frac{1}{2^7}$$ +$$\frac{1}{2^9}$$ } =q
then
$$q-\frac{q}{2}$$

$$\frac{q}{2}$$

now q is in GP and we need to find the sum of GP whose first term is $$\frac{1}{2}$$ and r is $$\frac{1}{4}$$ , n is 5
Either you can use the formula for sum of infinte terms of GP or genereal formula .

Lets try the general formula which says sum of n terms of GP

we have $$\frac{\frac{1}{2} * (1 - \frac{1}{4^5})}{ (1-\frac{1}{4})}$$

$$\frac{\frac{1}{2} * ( \frac{4^5-1}{4^5})}{ (\frac{3}{4})}$$

Can we say $$\frac{4^5-1}{4^5}$$ is nearly equal to 1

So this expression $$\frac{\frac{1}{2} * ( \frac{4^5-1}{4^5})}{ (\frac{3}{4})}$$ hence simplifies to $$\frac{2}{3}$$

now q= $$\frac{2}{3}$$

we need the value of $$\frac{q}{2}$$ = $$\frac{1}{3}$$

So Choice D suits our Answer.

PS: Does this solution look long - Yeah because i tried to explain each step. I am sure we will not being many steps that i wrote down . I did get my solution in 1: 03 min

abhimahna

I guess you missed on signs in this post , lucky you ended getting the right answer.

https://gmatclub.com/forum/kth-term-of- ... l#p1724824
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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10 Sep 2019, 20:19
prathns 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

Series has 10 terms where first term $$= \frac{1}{2}$$ and 10th term $$= - \frac{1}{2^{10}}$$.
Every odd is positive and every even term is negative.

The series becomes
$$\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - \frac{1}{64} + \frac{1}{128} - \frac{1}{256} + \frac{1}{512} - \frac{1}{1024}$$

Separating positive terms and negative terms we have
$$\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - \frac{1}{64} + \frac{1}{128} - \frac{1}{256} + \frac{1}{512} - \frac{1}{1024}$$
$$= \frac{1}{2}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) - \frac{1}{4}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..)$$
$$= (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) * (\frac{1}{2} - \frac{1}{4})$$
$$= \frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..)$$

And
$$1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} .. > 1.25$$
$$\frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) > \frac{1}{4} * 1.25$$
$$\frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) > 0.31$$

Checking answers only D is correct since $$\frac{1}{4} < 0.31 < \frac{1}{2}$$

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

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07 Oct 2019, 00:58
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Re: For every integer k from 1 to 10, inclusive, the kth term of   [#permalink] 07 Oct 2019, 00:58

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