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In the sequence al, a2, a3, • • • , aim the kthterm is defin

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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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89% (01:15) correct 10% (01:59) wrong based on 19 sessions


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) \(1\)
[Reveal] Spoiler: OA

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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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Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin   [#permalink] 11 Mar 2017, 20:57
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