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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
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(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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daviesj
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
.
.
\(S_{16} = 115\)approx
\(S_1 + S_2 + ...S_{16} = 100 + 101 + 102 + ... 115 = 16*100 + 15*16/2 = 1720\)
The sum will be a little more than 1720.
Answer (c)
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daviesj
\(S_1 = 100\)
\(S_2 = \frac{101^2 + 1}{101} \approx 101\) (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.
General Discussion
Posts: 3
Updated on: 19 Dec 2012, 21:53
Originally posted by muralilawson on 19 Dec 2012, 21:07.
Last edited by muralilawson on 19 Dec 2012, 21:53, edited 2 times in total.
Last edited by muralilawson on 19 Dec 2012, 21:53, edited 2 times in total.
VeritasPrepKarishma
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?
mbaiseasy
Joined: 13 Aug 2012
Last visit: 29 Dec 2013
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Concentration: Marketing, Finance
Schools: Full Time MBA (A$)
GPA: 3.23
Schools: Full Time MBA (A$)
Posts: 316
daviesj
S1 = \(100\)
S2 = \(100 + (1 + \frac{1}{101})\)
If you will notice \(1 + \frac{1}{101}\) is approximately 1... S2 = 100 + 1 is approx. ~ 101
S3 = \(101 + (1 + \frac{1}{101})\)
If you wil notice \(1 + \frac{1}{101}\) is approximately 1... S3 = 101 + 1 approx. ~ 102
Sum = 16 * 100 + 1 + 2 + 3 + ... + 15 = \(1600 + \frac{15(15+1)}{2} = 1600 + 120 = 1720\)
Answer: C
muralilawson
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.
daviesj
Answer: Option C
Please check the solution in attachment
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deep1624
100 + 101 + 102 + .... + 115
= 100 + (100 + 1) + (100 + 2) + (100 + 3) + .... (100 + 15)
= [100 + 100 + 100 + ... (16 times)] + [1 + 2 + 3 + ... 15]
= 100 * 16 + 15*16/2
The highlighted formula is that the sum of first n positive integers is given by n(n+1)/2.
So sum of first 15 positive integers is 15*16/2
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Kinshook
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daviesj
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?
\(S_1 = 100\)
\(S_2 = S_1 + 1 + \frac{1}{S_1 + 1} = 101 + \frac{1}{101} = 101.01 > 101\)
\(Q = S_1 + S_2 + .... + S_{16} = 100 + 101.01 + 102.01 + .... + 115.01 = \frac{16}{2} (100+115) + .01*15 = 8*215+.15 = 1720.15\)
IMO C
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?
\(S_1 = 100\)
\(S_2 = S_1 + 1 + \frac{1}{S_1 + 1} = 101 + \frac{1}{101} = 101.01 > 101\)
\(Q = S_1 + S_2 + .... + S_{16} = 100 + 101.01 + 102.01 + .... + 115.01 = \frac{16}{2} (100+115) + .01*15 = 8*215+.15 = 1720.15\)
IMO C
Akshaynandurkar
Joined: 29 Apr 2023
Last visit: 19 Jan 2025
Posts: 68
Location: India
GMAT Focus 1: 635 Q84 V86 DI74
GMAT Focus 1: 635 Q84 V86 DI74
Posts: 68
Write down the terms 1 by 1 to see the patterns
S1 = 100
S2 = 101 + 1/101
S3 = 102 + 1/101 + 1/(102+1/101)
S4 = 103 + 1/101 + 1/(102+1/101) + 1/(.....)
here as you can see the first term of each Sn term is significantly greater than each subsequent term and also terms after 1st is either 0 (for S1) or very close to 0
so there is no significant impact on summation of all terms of the terms after 1st
hence summation = 100+101+102+...115 = 16 x average of 1st and last terms (since this is AP)
summation = 16 * (100+115)/2 = 8 * 215 = 1720
hence the answer is somewhere between 1700 an 1750
S1 = 100
S2 = 101 + 1/101
S3 = 102 + 1/101 + 1/(102+1/101)
S4 = 103 + 1/101 + 1/(102+1/101) + 1/(.....)
here as you can see the first term of each Sn term is significantly greater than each subsequent term and also terms after 1st is either 0 (for S1) or very close to 0
so there is no significant impact on summation of all terms of the terms after 1st
hence summation = 100+101+102+...115 = 16 x average of 1st and last terms (since this is AP)
summation = 16 * (100+115)/2 = 8 * 215 = 1720
hence the answer is somewhere between 1700 an 1750
Posts: 257
S1=100, S2={100+1}+{1/101}=101+0.009, S3=102+0.009
Sum of S1..S16= 100,101,102,103...115
(Neglecting 0.009)
Sum S1...S16=Finding sum through Arithematic mean method
(115+100)/2= Mean of equally distanced numbers is (first+last)/2
(115+100)/2=(Sum S1...S16)/16
(Sum S1...S16)= 215*8=1720
Sum of S1..S16= 100,101,102,103...115
(Neglecting 0.009)
Sum S1...S16=Finding sum through Arithematic mean method
(115+100)/2= Mean of equally distanced numbers is (first+last)/2
(115+100)/2=(Sum S1...S16)/16
(Sum S1...S16)= 215*8=1720
daviesj
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