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# In the sequence a1, a2, a3,....a100, the kth term is defined as [m]ak=

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In the sequence a1, a2, a3,....a100, the kth term is defined as [m]ak=  [#permalink]

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31 Jan 2019, 21:21
00:00

Difficulty:

45% (medium)

Question Stats:

64% (01:29) correct 36% (03:01) wrong based on 13 sessions

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In the sequence a1, a2, a3,....a100, the kth term is defined as $$a_k= \frac{1}{k} - \frac{1}{k+1}$$ for all integers k from 1 through 100. What is the sum of 100 terms of the sequence?

A. $$\frac{1}{10100}$$

B. $$\frac{1}{100}$$

C. $$\frac{1}{101}$$

D. $$\frac{100}{101}$$

E. $$1$$

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Re: In the sequence a1, a2, a3,....a100, the kth term is defined as [m]ak=  [#permalink]

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31 Jan 2019, 22:00
Bunuel wrote:
In the sequence a1, a2, a3,....a100, the kth term is defined as $$a_k= \frac{1}{k} - \frac{1}{k+1}$$ for all integers k from 1 through 100. What is the sum of 100 terms of the sequence?

A. $$\frac{1}{10100}$$

B. $$\frac{1}{100}$$

C. $$\frac{1}{101}$$

D. $$\frac{100}{101}$$

E. $$1$$

$$a_k= \frac{1}{k} - \frac{1}{k+1}$$

$$a_1= \frac{1}{1} - \frac{1}{1+1}$$

$$a_2= \frac{1}{2} - \frac{1}{2+1}$$

$$a_100= \frac{1}{100} - \frac{1}{100+1}$$

Sum = (1/1) - (1/2) + (1/2) - (1/3) ...... (1/100+1/101 )

Sum = 1- (1/101)

Sum = 100/101

Hence D
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Re: In the sequence a1, a2, a3,....a100, the kth term is defined as [m]ak=   [#permalink] 31 Jan 2019, 22:00
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