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

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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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New post 23 Feb 2017, 01:16
cuhmoon wrote:
Hi Bunuel,

Can we say that if |r| <1, then it's an infinite GP ? How does one define infinite GP?
When does one use the forumula

Sum = b1 (r^n-1)/r-1 ?

Please explain..

Bunuel wrote:
Stiv wrote:
\(\frac{first \ term}{1-constant}\) Is this formula reversed when we have an increase by 0<constant<1? Does it look like this \(\frac{first \ term}{1+constant}\)?


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



To add to what Bunuel said, r is the common ratio and has nothing to do with whether a progression is infinite or finite. In either case, r can be less than 1 or more than 1

e.g.
1, 3, 9, 27, 81 ... (infinite sequence with r = 3)
27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3)
1, 3, 9 (finite sequence with 3 elements and r = 3)
27, 9, 3 (finite sequence with 3 elements and r = 1/3)

Now the point is that the sum of all terms of the first sequence is infinite. The terms will can getting larger and will keep adding up. So the sum will be infinite.

We can find the exact sum of the rest of the 3 sequences.

27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3)
Sum = a/(1 - r) = 27/(1 - 1/3) = 81/2

1, 3, 9 (finite sequence with 3 elements and r = 3)
Sum = a(r^n - 1)/(r - 1) = 1*(3^3 - 1)/(3 - 1) = 13

27, 9, 3 (finite sequence with 3 elements and r = 1/3)
Sum = a(1 - r^n)/(1 - r) = 27*(1 - 1/3^3)/(1 - 1/3) = 26*3/2 = 39
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Re: For every integer k from 1 to 10, inclusive the  [#permalink]

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New post 14 Jun 2017, 16:18
bekbek 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


We are given that for every integer k from 1 to 10 inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (1/2^k). We must determine the sum of the first 10 terms in the sequence. Before calculating the sum, we should recognize that the answer choices are provided as ranges of values, rather than as an exact value. Thus, we might not need to calculate the total of the 10 terms to determine an answer. Perhaps we can uncover a pattern to help us find the answer. Let’s start by listing out the first four terms.

k = 1:
(-1)^(1+1) x (1/2^1)
(-1)^2 x 1/2
1 x 1/2 = 1/2

k = 2:
(-1)^(2+1) x (1/2^2)
(-1)^3 x 1/4
-1 x 1/4 = -1/4

k = 3:
(-1)^(3+1) x (1/2^3)
(-1)^4 x 1/8
1 x 1/8 = 1/8

k = 4:
(-1)^(4+1) x (1/2^4)
(-1)^5 x 1/16
-1 x 1/16 = -1/16

Recall that we are trying to estimate the value of T = 1/2 + (-1/4) + 1/8 + (-1/16) + … until we have 10 terms. In other words, T = 1/2 – 1/4 + 1/8 – 1/16 + … until there are 10 terms.

We should notice that the absolute values of the terms are getting smaller:
|1/2|>|-1/4|>|1/8|>|-1/16|.

Notice that starting from the first term of 1/2, we are subtracting something less than 1/2 (notice that 1/4 < 1/2), but then adding back something even less (notice 1/8 < 1/4), and the process continues. Thus, because 1/2 and -1/4 are the largest term and the smallest term in our set, respectively, the sum will never fall below 1/4 or exceed 1/2.

Thus, we conclude that T is greater than 1/4 but less than 1/2.

Answer: D
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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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New post 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.

Answer: 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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New post 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
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Solution Kth Term.jpeg
Solution Kth Term.jpeg [ 24.64 KiB | Viewed 822 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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New post 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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New post 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 \(\frac{1}{4} < 0.31 < \frac{1}{2}\)

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

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