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# Sequence S is the sequence of numbers a1, a2, a3, ... , an.

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Founder
Joined: 18 Apr 2015
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Kudos [?]: 3169 [0], given: 11654

Sequence S is the sequence of numbers a1, a2, a3, ... , an. [#permalink]  31 Jul 2020, 04:49
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Question Stats:

55% (02:55) correct 44% (02:01) wrong based on 9 sessions
Sequence S is the sequence of numbers $$a_1, a_2, a_3,$$ ... , $$a_n$$. For each positive integer n, the $$n$$th number $$a_n$$ is defined by $$a_n=\frac{n+1}{3n}$$ . What is the product of the first 53 numbers in sequence S?

A. $$\frac{2}{3^{53}}$$

B. $$\frac{2}{3^{50}}$$

C. $$\frac{2}{3^{49}}$$

D. $$\frac{3}{2^{50}}$$

E. $$\frac{2}{3^{25}}$$
[Reveal] Spoiler: OA

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Founder
Joined: 18 Apr 2015
Posts: 12608
Followers: 268

Kudos [?]: 3169 [0], given: 11654

Re: Sequence S is the sequence of numbers a1, a2, a3, ... , an. [#permalink]  31 Jul 2020, 04:54
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Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
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Manager
Joined: 22 Jan 2020
Posts: 63
Followers: 0

Kudos [?]: 46 [1] , given: 8

Re: Sequence S is the sequence of numbers a1, a2, a3, ... , an. [#permalink]  03 Aug 2020, 04:33
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KUDOS
an=(n+1)/3n

a1=2/3
a2=3/6
a3=4/9
a4=5/12
and so on

the product of all 53 terms will look like

=(2/3)*(3*6)*(4/9)*(5/12)*....
=(2/(3*1))*(3/(3*2))*(4/(3*3))*(5/(3*4))...
Notice that each of the 53 terms in the denominator we have a 3 as a factor. So when all these 3's get multiplied together we get

(2*3*4*5*...53*54)/((3^53)*1*2*3*4...*53)
The terms 2 to 53 get cancelled out with ones in the numerator leaving
=54/(3^53)
=(27*2)/(3^53)
=((3^3)*2)/(3^53)
=2/(3^50)

Re: Sequence S is the sequence of numbers a1, a2, a3, ... , an.   [#permalink] 03 Aug 2020, 04:33
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