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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]  05 Mar 2017, 12:59
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Question Stats:

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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GRE 1: Q167 V156
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Kudos [?]: 1910 [3] , given: 397

Re: In the sequence al, a2, a3, • • • , aim the kthterm is defin [#permalink]  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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Sandy
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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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