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Intern  Joined: 05 Nov 2016
Posts: 1
Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 66% (02:23) correct 34% (02:42) wrong based on 52 sessions

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Sequence S is defined as $$S_n = X + \frac{1}{X}$$, where $$X = S_{n – 1} + 1$$, for all n > 1. If $$S_1= 201$$, then which of the following must be true of Q, the sum of the first 50 terms of S?

(A) 13,000 < Q < 14,000

(B) 12,000 < Q < 13,000

(C) 11,000 < Q < 12,000

(D) 10,000 < Q < 11,000

(E) 9,000 < Q < 10,000
Intern  B
Joined: 30 May 2017
Posts: 19
Re: Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1  [#permalink]

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S1 = 201

X= 201+1=202

S2=202+1/202 =+/-202

X = 202+1=203

S3=203+1/203 =+/-203

Q =+/- SUM(201+202+203+...+250) =+/- 11275 Answer C
Math Expert V
Joined: 02 Aug 2009
Posts: 8005
Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1  [#permalink]

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wittyeinstein wrote:
Sequence S is defined as $$S_n = X + \frac{1}{X}$$, where $$X = S_{n – 1} + 1$$, for all n > 1. If $$S_1= 201$$, then which of the following must be true of Q, the sum of the first 50 terms of S?

(A) 13,000 < Q < 14,000

(B) 12,000 < Q < 13,000

(C) 11,000 < Q < 12,000

(D) 10,000 < Q < 11,000

(E) 9,000 < Q < 10,000

$$S_n = X + \frac{1}{X}$$, where $$X = S_{n – 1} + 1$$
So 1/x = 1/201 is a very very small value as compared to x=201
So for such questions, I would just drop 1/x..
So
$$S_n = X = S_{n – 1} + 1$$, that is each term is ONE more than the previous term
So sum of first fifty terms is 201+202+203+.......250=200*50+1+2+3+4...+50=10000+(50*51)/2=10000+1275=11275 thus between 11000 and 12000

C
_________________ Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1   [#permalink] 04 Sep 2018, 04:23
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Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1

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