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655-705 (Hard)|   Sequences|      
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Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1 23 Jan 2020, 02:19
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65% (hard)

Question Stats:

65% (02:45) correct 35% (02:43) wrong based on 402 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
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C
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Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1 23 Jan 2020, 04:36
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Bunuel
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
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\(Q = S_1 + S_2 + S_3 + . . . . . . . + S_{50}\)

\(S_n = X + \frac{1}{X}\)
--> \(S_n = (S_{n-1} + 1) + \frac{1}{(S_{n-1} + 1)}\)

Given, \(S_1 = 201\)
--> \(S_2 = (S_1 + 1) + \frac{1}{(S_{n-1} + 1)} = 201 + 1 +\frac{ 1}{201 + 1} = 202 + \frac{1}{202} = 202\) (approx.)
Similarly \(S_3 = 203\), \(S_4 = 204\) . . . . . . . \(S_{50} = 250\)

--> Q = 201 + 202 + 203 + . . . . . . .. + 249 + 250 (approx.)
The above is in Arithmetic Progression with n = 50, first term = 201 & last term = 250

--> Sum of the terms, \(Q = \frac{50}{2}*(201 + 250) = 25*451 = 11275\)

Option C
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Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1 11 Feb 2021, 20:36
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VeritasKarishma plz explain this
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KarishmaB
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Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1 15 Feb 2021, 02:53
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Bunuel
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
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Given:
\(S_n = X + \frac{1}{X}\)
\(X = S_{n – 1} + 1\)

This means
\(X = S_{1} + 1 = 201 + 1=202\)
\(S_2 = 202 + \frac{1}{202}\)

So S2 is 1 more than S1 and then a tiny bit extra. S2 is approximately 1 more than S1. Then S3 will be approximately 1 more than S2 and so on. The only substantial increase from one term to the other is 1. The other increase is very very small (of the order of 1/200 which is .005. On the other hand, the options give a range of 1000. Hence, this tiny amount can be easily ignored.

First 50 terms will be
201, 202, 203 .. 250

Sum = Avg * 50 = (201 + 250)/2 * 50 = 451 * 25 = 11275

The actual sum will be a tiny bit more than this.

Answer (C)

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... gressions/
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Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1 21 Feb 2022, 17:31
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Bunuel
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
Show more

Breaking Down the Info:

X can be substituted, so write \(S_n = S_{n - 1} + 1 + \frac{1}{S_{n - 1} + 1}\)

We start with \(S_1 = 201\), then the fraction portion of each term, hence \(\frac{1}{S_{n - 1} + 1}\) , is negligible.

Then we can treat this sequence as \(S_n = S_{n-1} + 1\), which is an arithmetic sequence. The median of the 50 terms would be near \(225.5\). Then the sum is for our estimate is \(225.5*50 = 11250\). We expect the true value to be only a little bigger than that.

Answer: C
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Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1 06 Dec 2025, 00:48
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