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Difficulty: 605-655 Level,    Sequences,                                  
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
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SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So:
\(s_1=1-\frac{1}{2}\);
\(s_2=\frac{1}{2}-\frac{1}{3}\);
\(s_3=\frac{1}{3}-\frac{1}{4}\);
...

If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).

Question: is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)? --> is \(\frac{k}{k+1}>\frac{9}{10}\)? --> is \(k>9\)?

(1) k > 10. Sufficient.
(2) k < 19. Not sufficient.

Answer: A.
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
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(1) Take K as 11.
So, Sum = S1 + S2 + S3 + S4 + S5+ S6 + S7 + S8 + S9 + S10 + S11
Where,
S1 = 1 - (1/2)
S2 = (1/2) - (1/3)
S3 = (1/3) - (1/4) ...
S11 = (1/11) - (1/12)

Which implies, Sum = 1 - (1/12) /* The terms like +1/2, -1/2, +1/3, -1/3 will be added to zero. Only the first and last numbers remains */
==> Sum = 1 - 0.0XXXX > 9/10

If you take k as 12, the SUM = 1 - (1/13) which is again > 0.9 Hence SUFFICIENT

(2) K < 19
Consider K as 2. Then the sum is = 1 - (1/2) + (1/2) - (1/3) = 1 - (1/3) which is Less than 9/10
Consider K as 11, Then the sum is greater than 9/10 /* We already proved this in (1) above */
Hence (2) is In Sufficient
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
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Bunuel wrote:
The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?


(1) k > 10

(2) k < 19


Data Sufficiency
Question: 107
Category: Arithmetic Sequences
Page: 160
Difficulty: 650


The Official Guide For GMAT® Quantitative Review, 2ND Edition


Similar question to practice: https://gmatclub.com/forum/the-numbers- ... 06213.html
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
Bunuel wrote:
SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So:
\(s_1=1-\frac{1}{2}\);
\(s_2=\frac{1}{2}-\frac{1}{3}\);
\(s_3=\frac{1}{3}-\frac{1}{4}\);
...

If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).

Question: is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)? --> is \(\frac{k}{k+1}>\frac{9}{10}\)? --> is \(k>9\)?

(1) k > 10. Sufficient.
(2) k < 19. Not sufficient.

Answer: A.



Hi Bunuel

I understood your point but I approached the problem in the following way :

Point 1: I saw the word “Sequence” — that’s why I have written the formula of Sum of n integers as :
Sum of first n integers in s sequence =
n/2( 1st term + Last term )

We have to prove that : Sum of 1st K terms in the sequence is > 9/10 i.e greater than 0.9

1) as per the first statement , K>10

So I took k’s value as 11 to check whether the sun is greater than 0.9 or not ,

Sum (11 integers ) = 11/2{ 1/2 + [(1/11)-1/12] }
Which gives something around 5.6
So , it’s not true

Again I took k =100
Sum (100 integers ) =100/2{1/2 + [1/100-1/101]}
The answer is something around 55
Now it’s true

That’s why A is not sufficient

Plz guide me regarding this approach for which I was pretty confident initially :(

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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
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LeenaSai wrote:
Bunuel wrote:
SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So:
\(s_1=1-\frac{1}{2}\);
\(s_2=\frac{1}{2}-\frac{1}{3}\);
\(s_3=\frac{1}{3}-\frac{1}{4}\);
...

If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).

Question: is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)? --> is \(\frac{k}{k+1}>\frac{9}{10}\)? --> is \(k>9\)?

(1) k > 10. Sufficient.
(2) k < 19. Not sufficient.

Answer: A.



Hi Bunuel

I understood your point but I approached the problem in the following way :

Point 1: I saw the word “Sequence” — that’s why I have written the formula of Sum of n integers as :
Sum of first n integers in s sequence =
n/2( 1st term + Last term )

We have to prove that : Sum of 1st K terms in the sequence is > 9/10 i.e greater than 0.9

1) as per the first statement , K>10

So I took k’s value as 11 to check whether the sun is greater than 0.9 or not ,

Sum (11 integers ) = 11/2{ 1/2 - [(1/11)-1/12] }
Which gives something around 5.6
So , it’s not true

Again I took k =100
Sum (100 integers ) =100/2{1/2-[1/100-1/101]}
The answer is something around 55
Now it’s true

That’s why A is not sufficient

Plz guide me regarding this approach for which I was pretty confident initially :(

Posted from my mobile device


The formula you quote, is for evenly spaced sets (aka arithmetical progression), and cannot be applied to all sequences. The sequence at hand is 1/2, 1/6, 1/12, 1/20, ... As you can see this is not an arithmetic progression, so you cannot use the formula you used.

Check links below for more.


12. Sequences



For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
avigutman wrote:
Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1


Hi
I don’t understand when we put k in expression Sn we solve to get 1/k(k+1)


We have to take LCM and cancel the rest. Which will result in 1/k - 1/k+1 = 1/k(1+k)


Correct me and help me with correct reasoning here please.

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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
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iamvishnu wrote:
I don’t understand when we put k in expression Sn we solve to get 1/k(k+1)


We have to take LCM and cancel the rest. Which will result in 1/k - 1/k+1 = 1/k(1+k)


Correct me and help me with correct reasoning here please.


Yes, your algebra is correct iamvishnu. But that's not a way to solve this problem in 2 minutes, IMHO.
Plugging in n=1, then n=2, etc (keeping the expression as it was, without doing the algebra) shows a pattern in which consecutive terms cancel one another, as I show in the video.
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
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Bunuel wrote:
SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So:
    \(s_1=1-\frac{1}{2}\);
    \(s_2=\frac{1}{2}-\frac{1}{3}\);
    \(s_3=\frac{1}{3}-\frac{1}{4}\);
    ...

If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).

Question:
    Is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)?
    Is \(\frac{k}{k+1}>\frac{9}{10}\)?
    Is \(k>9\)?


(1) k > 10. Sufficient.
(2) k < 19. Not sufficient.

Answer: A.



Hi Bunuel

Could you please explain to me the move highlighted in red? How did you move from is k/k+1>9/10 to is k>9?
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
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HoudaSR wrote:
Bunuel wrote:
SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So:
    \(s_1=1-\frac{1}{2}\);
    \(s_2=\frac{1}{2}-\frac{1}{3}\);
    \(s_3=\frac{1}{3}-\frac{1}{4}\);
    ...

If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).

Question:
    Is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)?
    Is \(\frac{k}{k+1}>\frac{9}{10}\)?
    Is \(k>9\)?


(1) k > 10. Sufficient.
(2) k < 19. Not sufficient.

Answer: A.



Hi Bunuel

Could you please explain to me the move highlighted in red? How did you move from is k/k+1>9/10 to is k>9?


    \(1-\frac{1}{k+1}>\frac{9}{10}\);

    \(\frac{(k+1) - 1}{k+1}>\frac{9}{10}\);

    \(\frac{k}{k+1}>\frac{9}{10}\);

    \(10k > 9(k+1)\);

    \(k>9\).

Hope it helps.
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) [#permalink]
Bunuel wrote:
HoudaSR wrote:
Bunuel wrote:
SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So:
    \(s_1=1-\frac{1}{2}\);
    \(s_2=\frac{1}{2}-\frac{1}{3}\);
    \(s_3=\frac{1}{3}-\frac{1}{4}\);
    ...

If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).

Question:
    Is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)?
    Is \(\frac{k}{k+1}>\frac{9}{10}\)?
    Is \(k>9\)?


(1) k > 10. Sufficient.
(2) k < 19. Not sufficient.

Answer: A.



Hi Bunuel

Could you please explain to me the move highlighted in red? How did you move from is k/k+1>9/10 to is k>9?


    \(1-\frac{1}{k+1}>\frac{9}{10}\);

    \(\frac{(k+1) - 1}{k+1}>\frac{9}{10}\);

    \(\frac{k}{k+1}>\frac{9}{10}\);

    \(10k > 9(k+1)\);

    \(k>9\).

Hope it helps.


It is clear. Thank you Bunuel for your prompt response.
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