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In the sequence al, a2, a3, • • • , aim the kthterm is defin
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Carcass
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In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  05 Mar 2017, 12:59
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In the sequence \(a_1\), \(a_2\), \(a_3\), • • • ,\(a_{100}\), the \(k_{th}\) term is defined by \(a_k = \frac{1}{k} - \frac{1}{k+1}\) for all integers \(k\) from 1 through 100. What is the sum of the 100 terms of this sequence?

A) \(\frac{1}{10,100}\)

B) \(\frac{1}{101}\)

C) \(\frac{1}{100}\)

D) \(\frac{100}{101}\)

E) \(1\)
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sandy
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  11 Mar 2017, 20:57
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Explanation

Remember, the key to most GRE sequence questions is discovering a pattern among the terms. So start off by finding the first few terms and looking for a pattern.

\(a1 = 1 -\frac{1}{1+1}= 1-\frac{1}{2}\).
\(a2 = \frac{1}{2} -\frac{1}{2+1}= \frac{1}{2} - \frac{1}{3}\).
\(a3 = \frac{1}{3} -\frac{1}{3+1}= \frac{1}{3} - \frac{1}{4}\).

Hence sum \(a1 + a2 + a3 ....... = 1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ......\)

So alternate terms terms cancel out. So

\(a100 = \frac{1}{100} - \frac{1}{101}\)

So \(a1 + a2 + a3 ...... a100\)=\(1 - \frac{1}{101} = \frac{100}{101}\).


Hence option D is correct.
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as005
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  13 Mar 2021, 13:05
Carcass wrote:


In the sequence \(a_1\), \(a_2\), \(a_3\), • • • ,\(a_{100}\), the \(k_{th}\) term is defined by \(a_k = \frac{1}{k} - \frac{1}{k+1}\) for all integers \(k\) from 1 through 100. What is the sum of the 100 terms of this sequence?

A) \(\frac{1}{10,100}\)

B) \(\frac{1}{101}\)

C) \(\frac{1}{100}\)

D) \(\frac{100}{101}\)

E) \(1\)


Why don't we get the correct answer for this using the sum of n terms of AP formula?

I tried: \(K1 = 0.5; K100= 0.000099; Average of K1 and K100 = 0.2500495; No. of terms in AP = 100; So, sum should be 25.\)
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KarunMendiratta
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  13 Mar 2021, 21:57
as005 wrote:
Carcass wrote:


In the sequence \(a_1\), \(a_2\), \(a_3\), • • • ,\(a_{100}\), the \(k_{th}\) term is defined by \(a_k = \frac{1}{k} - \frac{1}{k+1}\) for all integers \(k\) from 1 through 100. What is the sum of the 100 terms of this sequence?

A) \(\frac{1}{10,100}\)

B) \(\frac{1}{101}\)

C) \(\frac{1}{100}\)

D) \(\frac{100}{101}\)

E) \(1\)


Why don't we get the correct answer for this using the sum of n terms of AP formula?

I tried: \(K1 = 0.5; K100= 0.000099; Average of K1 and K100 = 0.2500495; No. of terms in AP = 100; So, sum should be 25.\)


They are not in A.P dear!

\(a_1 = \frac{1}{2}\)

\(a_2 = \frac{1}{6}\)

\(a_3 = \frac{1}{12}\)

Common difference (d) = \(a_2 - a_1 = a_3 - a_2\)

\(a_2 - a_1 = \frac{-1}{3}\)
\(a_3 - a_2 = \frac{-1}{12}\)
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as005
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  14 Mar 2021, 02:02
KarunMendiratta wrote:
as005 wrote:
Carcass wrote:


In the sequence \(a_1\), \(a_2\), \(a_3\), • • • ,\(a_{100}\), the \(k_{th}\) term is defined by \(a_k = \frac{1}{k} - \frac{1}{k+1}\) for all integers \(k\) from 1 through 100. What is the sum of the 100 terms of this sequence?

A) \(\frac{1}{10,100}\)

B) \(\frac{1}{101}\)

C) \(\frac{1}{100}\)

D) \(\frac{100}{101}\)

E) \(1\)


Why don't we get the correct answer for this using the sum of n terms of AP formula?

I tried: \(K1 = 0.5; K100= 0.000099; Average of K1 and K100 = 0.2500495; No. of terms in AP = 100; So, sum should be 25.\)


They are not in A.P dear!

\(a_1 = \frac{1}{2}\)

\(a_2 = \frac{1}{6}\)

\(a_3 = \frac{1}{12}\)

Common difference (d) = \(a_2 - a_1 = a_3 - a_2\)

\(a_2 - a_1 = \frac{-1}{3}\)
\(a_3 - a_2 = \frac{-1}{12}\)


Thank you! What kind of progression would this be called?
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KarunMendiratta
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In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  14 Mar 2021, 02:08
as005 wrote:

Thank you! What kind of progression would this be called?


It is called as Telescoping Series
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as005
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  14 Mar 2021, 02:12
KarunMendiratta wrote:
as005 wrote:

Thank you! What kind of progression would this be called?


It is called as Telescoping Series

Wow. Every day learning something new. Thanks for your help! :)
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Carcass
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  14 Mar 2021, 02:46
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KarunMendiratta wrote:
as005 wrote:

Thank you! What kind of progression would this be called?


It is called as Telescoping Series


it is relevant to know for the GRE ?

I mean the question could be solved without knowing this
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KarunMendiratta
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  14 Mar 2021, 02:48
Carcass wrote:
KarunMendiratta wrote:
as005 wrote:

Thank you! What kind of progression would this be called?


It is called as Telescoping Series


it is relevant to know for the GRE ?

I mean the question could be solved without knowing this


Totally, we don't need to know it
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I hope this helps!

Regards:

Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, India
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Kayto
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  20 Feb 2022, 09:16
KarunMendiratta wrote:
as005 wrote:
Carcass wrote:


In the sequence \(a_1\), \(a_2\), \(a_3\), • • • ,\(a_{100}\), the \(k_{th}\) term is defined by \(a_k = \frac{1}{k} - \frac{1}{k+1}\) for all integers \(k\) from 1 through 100. What is the sum of the 100 terms of this sequence?

A) \(\frac{1}{10,100}\)

B) \(\frac{1}{101}\)

C) \(\frac{1}{100}\)

D) \(\frac{100}{101}\)

E) \(1\)


Why don't we get the correct answer for this using the sum of n terms of AP formula?

I tried: \(K1 = 0.5; K100= 0.000099; Average of K1 and K100 = 0.2500495; No. of terms in AP = 100; So, sum should be 25.\)


They are not in A.P dear!

\(a_1 = \frac{1}{2}\)

\(a_2 = \frac{1}{6}\)

\(a_3 = \frac{1}{12}\)

Common difference (d) = \(a_2 - a_1 = a_3 - a_2\)

\(a_2 - a_1 = \frac{-1}{3}\)
\(a_3 - a_2 = \frac{-1}{12}\)


Out of curiosity what does AP mean. Does that have to do with the set being evenly spaced? So I am assuming you can only use the formula sum = n (a1+a2/2) when the set is evenly spaced?
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Carcass
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  20 Feb 2022, 10:16
Expert Reply
arithmetic Progression

Look at here https://greprepclub.com/forum/gre-math- ... tml#p81701 and here https://greprepclub.com/forum/gre-math- ... html#p5077
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GMATGuruNY
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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink] New post  20 Feb 2022, 14:19
1
Carcass wrote:


In the sequence \(a_1\), \(a_2\), \(a_3\), • • • ,\(a_{100}\), the \(k_{th}\) term is defined by \(a_k = \frac{1}{k} - \frac{1}{k+1}\) for all integers \(k\) from 1 through 100. What is the sum of the 100 terms of this sequence?

A) \(\frac{1}{10,100}\)

B) \(\frac{1}{101}\)

C) \(\frac{1}{100}\)

D) \(\frac{100}{101}\)

E) \(1\)


Alternate approach:
Calculate smaller sums and LOOK FOR A PATTERN.

First term = 1 - 1/2 = 1/2
Second term = 1/2 - 1/3 = 1/6
Third term = 1/3 - 1/4 = 1/12

If the sequence includes only 1 term, the sum = 1/2
If the sequence includes only the first 2 terms, the sum = 1/2 + 1/6 = 2/3
If the sequence includes only the first 3 terms, the sum = 2/3 + 1/12 = 3/4

Notice the pattern:
The numerator of the sum = the number of terms
The denominator of the sum = 1 more than the numerator

Thus:
Sum of 100 terms = 100/101

Show: ::
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