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Status: Never ever give up on yourself.Period.
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Sequence S is defined as Sn=Sn-1 + 1 +1/(Sn-1 + 1) for all n [#permalink]
 19 Dec 2012, 07:18
Question Stats:
43% (03:55) correct
56% (00:41) wrong based on 30 sessions
Sequence S is defined as S n = (S n-1 +1) + {1 / (S n-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S? (A) 1,600 ≤ Q ≤ 1,650 (B) 1,650 ≤ Q ≤ 1,700 (C) 1,700 ≤ Q ≤ 1,750 (D) 1,750 ≤ Q ≤ 1,800 (E) 1,800 ≤ Q ≤ 1,850
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Last edited by daviesj on 20 Dec 2012, 00:09, edited 1 time in total.
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Director
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Re: Sequence S is defined as for all n > 121. [#permalink]
19 Dec 2012, 08:49
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daviesj wrote: Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850
method to solve plz... S_1 = 100S_2 = \frac{101^2 + 1}{101} \approx 101 (Since 1 is negligible when compared to 101^2) So, the series is almost an arithmetic progression with a=100, d=1, We have got to find the sum of "n" terms where "n" is 16. S_{16} = \frac{16}{2}*(2*100 + (16-1)*1) = 8*215 = 1720 Answer is C.
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Re: Sequence S is defined as for all n > 121. [#permalink]
19 Dec 2012, 21:14
daviesj wrote: Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850
method to solve plz... S1 has been given a big value i.e. 100 instead of the usual 0/1 etc. Why? Because 1/100 is negligible when added to 101 S_n = S_{n-1} + 1 + \frac{1}{S_{n-1}}S_2 = S_{1} + 1 + \frac{1}{S_{1}} = 100 + 1 + \frac{1}{100} = 101 approx S_3 = 101+ 1 + 1/101 = 102 approx . . S_{16} = 115approx S_1 + S_2 + ...S_{16} = 100 + 101 + 102 + ... 115 = 16*100 + 15*16/2 = 1720The sum will be a little more than 1720. Answer (c)
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Re: Sequence S is defined as for all n > 121. [#permalink]
19 Dec 2012, 22:07
VeritasPrepKarishma wrote: daviesj wrote: Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850
method to solve plz... S1 has been given a big value i.e. 100 instead of the usual 0/1 etc. Why? Because 1/100 is negligible when added to 101 S_n = S_{n-1} + 1 + \frac{1}{S_{n-1}}S_2 = S_{1} + 1 + \frac{1}{S_{1}} = 100 + 1 + \frac{1}{100} = 101 approx S_3 = 101+ 1 + 1/101 = 102 approx . . S_{16} = 115approx S_1 + S_2 + ...S_{16} = 100 + 101 + 102 + ... 115 = 16*100 + 15*16/2 = 1720The sum will be a little more than 1720. Answer (c) i would say the question is wrong or, atleast, not an exact GMAT type question... why to assume N as an integer.. .it is not specified in the question that n is an integer....n could be 1.2, 1.2,....etc for n>1 when n is not an integer.... i am thinking in the GMAT prospective...what is the source of this qtn?
Last edited by muralilawson on 19 Dec 2012, 22:53, edited 2 times in total.
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Senior Manager
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Re: Sequence S is defined as for all n > 121. [#permalink]
19 Dec 2012, 22:19
daviesj wrote: Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850
method to solve plz... S1 = 100S2 = 100 + (1 + \frac{1}{101}) If you will notice 1 + \frac{1}{101} is approximately 1... S2 = 100 + 1 is approx. ~ 101S3 = 101 + (1 + \frac{1}{101})If you wil notice 1 + \frac{1}{101} is approximately 1... S3 = 101 + 1 approx. ~ 102Sum = 16 * 100 + 1 + 2 + 3 + ... + 15 = 1600 + \frac{15(15+1)}{2} = 1600 + 120 = 1720Answer: C
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Director
Joined: 02 Jul 2012
Posts: 790
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Followers: 19
Kudos [?]:
277
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Re: Sequence S is defined as for all n > 121. [#permalink]
19 Dec 2012, 22:52
muralilawson wrote: VeritasPrepKarishma wrote: daviesj wrote: Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850
method to solve plz... S1 has been given a big value i.e. 100 instead of the usual 0/1 etc. Why? Because 1/100 is negligible when added to 101 S_n = S_{n-1} + 1 + \frac{1}{S_{n-1}}S_2 = S_{1} + 1 + \frac{1}{S_{1}} = 100 + 1 + \frac{1}{100} = 101 approx S_3 = 101+ 1 + 1/101 = 102 approx . . S_{16} = 115approx S_1 + S_2 + ...S_{16} = 100 + 101 + 102 + ... 115 = 16*100 + 15*16/2 = 1720The sum will be a little more than 1720. Answer (c) i would say the question is wrong or, atleast, not an exact GMAT question... why to assume N as an integer.. .it is not specified in the question that n is an integer....n could be 1.2, 1.2,....etc for n>1 when n is not an integer.... i am thinking in the GMAT prospective... Since this is a PS question and not a DS question, we are free to make that assumption. "n" is only a subscript indicating the ordinal number of each term and hence can be taken to be integers.
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Re: Sequence S is defined as for all n > 121.
[#permalink]
19 Dec 2012, 22:52
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