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23 Jan 2020, 02:19
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C
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Difficulty:
65%
(hard)
Question Stats:
66% (02:45) correct
34%
(02:43)
wrong
based on 403
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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
(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
23 Jan 2020, 04:36
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Bunuel
\(Q = S_1 + S_2 + S_3 + . . . . . . . + S_{50}\)
\(S_n = X + \frac{1}{X}\)
--> \(S_n = (S_{n-1} + 1) + \frac{1}{(S_{n-1} + 1)}\)
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
General Discussion
11 Feb 2021, 20:36
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VeritasKarishma plz explain this
