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Veritas Prep 10 Year Anniversary Promo Question #3

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Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:00
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Veritas Prep 10 Year Anniversary Promo Question #3


One quant and one verbal question will be posted each day starting on Monday Sept 17th at 10 AM PST/1 PM EST and the first person to correctly answer the question and show how they arrived at the answer will win a free Veritas Prep GMAT course ($1,650 value). Winners will be selected and notified by a GMAT Club moderator. For more questions and details please check here: veritas-prep-10-year-anniversary-giveaway-138806.html

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A sequence is given by the rule a_{n} = |a_{(n-2)}| - |a_{(n-1)}| for all n\geq{3}, where a_1 = 0 and a_2 = 3.

A function s_{n} is defined as the sum of all the terms of the sequence from its beginning through a_n. For instance, s_{4} = a_1 + a_2 + a_3 + a_4. What is s_{101}?


(A) -3
(B) 0
(C) 3
(D) 201
(E) 303
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:00
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Official Explanation:

Answer is C

While we spend a lot of time honing the skill of translating English into algebra, there is sometimes great comfort to be gained in a sequence problem through doing precisely the reverse. The general rule for this sequence is that we derive each term based on the two terms that precede it – specifically by subtracting the absolute value of the previous term from the absolute value of the “pre-previous” one.

So a_3 = |a_1| - |a_2|, and a_4 = |a_2| - |a_3|, and a_5 = |a_3| - |a_4|, and so on. We are told that the sequence begins 0, 3, ..., so we can derive that:

a_3 = |a_1| - |a_2| = |0| - |3| = 0 - 3 = -3.
Then a_4 = |a_2| - |a_3| = |3| - |-3| = 3 - 3 = 0.
Then a_5 = |a_3| - |a_4| = |-3| - |0| = 3 - 0 = 3.

As soon as we’ve seen a_4 and a_5 turn out to be 0 and 3 consecutively, we know that the next number to show up in the sequence will be identical to the number that showed up after the last time we saw 0 and 3 appear consecutively (i.e. as a_1 and a_2) — so a_6 will be identical to a_3, so -3. So we have established the pattern 0, 3, -3, 0, 3, -3, ... for our sequence.

Every time we finish a complete cycle within the sequence, the sum returns to 0 (since 0 + 3 + -3 = 0). We finish a cycle after every third entry (i.e. after the third, the sixth, the ninth, and so on), so we will have done so (and returned our running sum to 0) with a_{99}. a_{100} will then add itself (0) to that sum, and a_{101} will add itself (3) onto that. So we will land at a sum of 3 for s_{101}.
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:05
Answer is E as the sum of any first two terms is 3 & then 6, 6, 12, 12 & so on
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:08
first 8 terms 0,3,-3,0,3,-3,0,3

sum of every four terms is 0+3-3+0 = 0 sum of S100 = 0 sum of s101 = 0+3 =3

Answer is 3 (C)
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:10
Ans is C,

This would be a series of -3 , 0, 3,0 ... .
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Last edited by Vips0000 on 18 Sep 2012, 09:40, edited 1 time in total.
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:12
ANswer C.


S101 = a1+ a2+ .....+ a101

a3=|a1|-|a2|= -3
a4=0

Follwoing the pattern,,

0,3,-3,0,3,-3,....

S99 = 0

A100=0

A101=3

Hence,S101=3
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:14
Answer is C

a = {0,3,-3,0,3,-3,0,3,-3....} repeated every 3 counts
s101 = contains 33 triplets + 2 so the sum is 33 * (0+3-3) + (0+3) = 3

Last edited by jirayr on 18 Sep 2012, 09:35, edited 1 time in total.
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:18
IMo - C

pattern
0,3,-3, ...
sum of 3 terms = 0
sum of 99 terms = 0
100 th = 0
101st = 3

sum = 3
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:47
Ans=C,
i have considered first upto a12, and observed that s1=0, s2=3, s3=0 and the cycle repeats..
by this we can divide 101 as 99+2 i.e., s99 would be 0 and s100 =0 and s101=3.

Last edited by chetanachethu on 18 Sep 2012, 10:05, edited 2 times in total.
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 09:51
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Considering the rule of Cyclicity,we can conclude that the pattern repeat itself after 3 units i.e a1=0,a2=3,a3=-3 ; again a4=0,a5=3,a6=-3 and so on...

Now 101/3 [as the pattern repeat itself after 3 units as above] gibes a quotient 99 and thus sum of a1+a2+..a99=0 [as every 3 units having sum 0]and a100+a101=|0|+|3|=3

So, S101=a1+a2+..a99+a100+a101=3.

So Answer is 3.
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 10:24
Answer : C

We have
an = |a(n-2)| - |a(n-1)|
a1 = 0
a2 =3

If we use the above formula we get a repetitve series of 3,0,-3 as follows
a3 = |a2|- |a1| = 3-0 = 3
a4 = |a3| - |a2| = 3-3 =0
a5 = |a4|-|a3| = 0 - 3 = -3
a6 = |a5| - |a4| = |-3| - 0 = 3-0 =3
a7 = |a6| - |a5| = |3| - |-3| = 3 -3 = 0
And this trend will continue as follows a8 = -3 , a9 = 3 ....and so on until a101 = -3
If we observe there is a pattern emerging starting a3 .If we sum a particular triplet,say a3 + a4+ a5 = 3+0+-3 = 0.
Our sum will be 0
Similiarly if we keep adding triplets
a6 + a7 + a8 =0
a9 + a10 + a11 = 0
.
.
.
a99 + a100 + a101 = 3 + 0 + -3 = 0.
So we can conclude the sum of all the triplets is 0.
a3 + .a4 + a5 + .......a101 = 0.

S101 = a1 + a2 + (sum of all the triplets present )
= 0 + 3 + 0
Hence S101 = 3.

Answer C
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 10:38
Bunuel wrote:
A sequence is given by the rule a_{n} = |a_{(n-2)}| - |a_{(n-1)}| for all n\geq{3}, where a_1 = 0 and a_2 = 3.

A function s_{n} is defined as the sum of all the terms of the sequence from its beginning through a_n. For instance, s_{4} = a_1 + a_2 + a_3 + a_4. What is s_{101}?


(A) -3
(B) 0
(C) 3
(D) 201
(E) 303



a3 = |a1| - |a2|
a4 = |a2| - |a3|
a5 = |a3| - |a4|
..
..
..
..
a100 = |a98| - |a99|
a101 = |a99| - |a100|
----------------------
Summing together both sides
a3+a4+a5+.......+a101 = |a1| - |a100| ( because all other values will be cancelled out in right hand side. For example, |a2| from 1st equation will be cancelled out from |a2| from 2nd equation, and so on.)

adding a1 + a2 both sides
a1 + a2 +a3 .............. + a101 = S101 = a1+a2+|a1| - |a100|
=> 0 + 3 + |0| - |a100|

Now these terms follow a pattern as follows:
a3 = -3
a4 = 0
a5 = -3
..
..
..
So, EVEN indexed "a" will be zero.
Thus, => 0+3+|0|+0 = 3

Hence, answer is C.
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 10:56
a1=0, a2=3, a3=-3, a4=0, a5=3, a6=-3.......
repetition occurs after every 3rd unit/count.
S101=a1+a2+a3....+a99+a100+a101
s101=0+a100+a101 as [101][/3] leaves a remainder 2; therefore sum (a1+a2+a3+...a99)=0
S101=0+0+3
=3, (c)
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 12:08
So I figured out the sequence first
a1=0, a2=3, a3=-3, a4=0, a5=3, a6=-3, a7=0, a8=3, a9=-3, a10=0.......
Every third term is a -3 for a3, a6, a9, a12, a15, a18, a21, a, 24....... They are all -3.
So s102 = -3 therefore s101 = 3
Another is that the sum of the first 10 is 0 s10= s1 + s2 +s3 + s4....... is 0 so every tenth interval is 0, therefore s100=0
The answer is 3 which is option C
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 12:46
A sequence is given by the rule a_{n} = |a_{(n-2)}| - |a_{(n-1)}| for all n\geq{3}, where a_1 = 0 and a_2 = 3.

A function s_{n} is defined as the sum of all the terms of the sequence from its beginning through a_n. For instance, s_{4} = a_1 + a_2 + a_3 + a_4. What is s_{101}?[/b]

(A) -3
(B) 0
(C) 3
(D) 201
(E) 303

Answer is C:

Explaination:
a(1) = 0 (given)
a(2) = 3 (given)
a(3) = |a(1)| - |a(2)| = |0| - |3| = 0 - 3 = -3
a(4) = |a(2)| - |a(3)| = |3| - |-3| = 3 - 3 = 0
a(5) = |a(3)| - |a(4)| = |-3| - |0| = 3 - 0 = 3
a(6) = |a(4)| - |a(5)| = |0| - |3| = 0 - 3 = -3

So the series is 0, 3, -3, 0, 3, -3, for a(1), a(2), a(3), a(4), a(5) and a(6) respectively. If you look at the series carefully, S(3) = 0 + 3 + (-3) = 0
this trend will continue and hence S(3), S(6), S(9)..... S(99) = 0
S(101) = S(99) + a(100) + a(101) = 0 + 0 + 3 = 3
S(101) = 3


OA please?
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 13:06
Answer is C!

The pattern is 0,3,-3,0,3,-3,0,3,-3...

101/3=99 R2

0+3=3
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 17:12
The answer is (C)3

Solution:
We have: S101 =a1+ a2+a3+a4…..+a99+a100+a101
= a1+ a2+|a1 |-|a2 |+|a2 |-|a3 |+⋯.+|a97 |-|a98 |+|a98 |-|a99 |+|a99 |-|a100 |
= a1+ a2+|a1 |-|a100 |
= 3-|a100 | (1)

Then: a1=0; a2=3; a3=|a1 |-|a2 |=0-3=-3; a4=|a2 |-|a3 |=3-3=0; a5=|a3 |-|a4 |=3-0=3;
Similarly: a4=0; a5=3; a6=|a4 |-|a5 |=0-3=-3; a7=|a_5 |-|a6 |=3-3=0; a8=|a6 |-|a7 |=3-0=3;
From those results,we can conclude that: a(3n+1)=0;a(3n+2)=3;a(3n)=-3

So, a_100=a(3×33+1)=0 (2)
From (1) and (2), we have: S101= 3-|a100 | = 3
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 18:29
Answer is C:

Explaination:
a(1) = 0 (given)
a(2) = 3 (given)
a(3) = |a(1)| - |a(2)| = |0| - |3| = 0 - 3 = -3
a(4) = |a(2)| - |a(3)| = |3| - |-3| = 3 - 3 = 0
a(5) = |a(3)| - |a(4)| = |-3| - |0| = 3 - 0 = 3
a(6) = |a(4)| - |a(5)| = |0| - |3| = 0 - 3 = -3

However the question asks for S(101). So we have to find a series for S(n).
S(1) = a(1) = 0
S(2) = a(1) + a(2) = 0 + 3 = 3
S(3) = a(1) + a(2) + a(3) = 0 + 3 - 3 = 0
S(4) = a(1) + a(2) + a(3) + a(4) = 0 + 3 - 3 + 0 = 0
s(5) = a(1) + a(2) + a(3) + a(4) + a(5) = 0 + 3 - 3 + 0 + 3 = 3
s(6) = a(1) + a(2) + a(3) + a(4) + a(5) +a(6) = 0 + 3 - 3 + 0 + 3 - 3 = 0

So the real series is 0, 3, 0, 0, 3, 0, for s(1), s(2), s(3), s(4), s(5) and s(6) respectively.

If you divide 101/3 you get 33 remainder 2, which tells us the correct answer should be the 2nd term of the series,
hence S(101) = 3
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 18 Sep 2012, 21:23

Solution


As per the question we have values of terms
a1 = 0;a2 =3;a3 = -3

Using the function a_{n} = |a_{(n-2)}| - |a_{(n-1)}| for all n\geq{3}, can derive

a4 = 0

If we continue solving the next two terms (i.e a5 & a6) we will see that these terms are cyclic i.e (0,3,-3)

Moving forward we have,

s_{4} = a_1 + a_2 + a_3 + a_4

Calculating the first four terms s (4) = 0 ; In the same way s(100) = 0 and s(101) = s(100) + a101 = 3

Hence OA is C
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Re: Veritas Prep 10 Year Anniversary Promo Question #3 [#permalink] New post 19 Sep 2012, 08:01
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We have a winner!



Winner:

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Official Explanation:

Answer is C

While we spend a lot of time honing the skill of translating English into algebra, there is sometimes great comfort to be gained in a sequence problem through doing precisely the reverse. The general rule for this sequence is that we derive each term based on the two terms that precede it – specifically by subtracting the absolute value of the previous term from the absolute value of the “pre-previous” one.

So a_3 = |a_1| - |a_2|, and a_4 = |a_2| - |a_3|, and a_5 = |a_3| - |a_4|, and so on. We are told that the sequence begins 0, 3, ..., so we can derive that:

a_3 = |a_1| - |a_2| = |0| - |3| = 0 - 3 = -3.
Then a_4 = |a_2| - |a_3| = |3| - |-3| = 3 - 3 = 0.
Then a_5 = |a_3| - |a_4| = |-3| - |0| = 3 - 0 = 3.

As soon as we’ve seen a_4 and a_5 turn out to be 0 and 3 consecutively, we know that the next number to show up in the sequence will be identical to the number that showed up after the last time we saw 0 and 3 appear consecutively (i.e. as a_1 and a_2) — so a_6 will be identical to a_3, so -3. So we have established the pattern 0, 3, -3, 0, 3, -3, ... for our sequence.

Every time we finish a complete cycle within the sequence, the sum returns to 0 (since 0 + 3 + -3 = 0). We finish a cycle after every third entry (i.e. after the third, the sixth, the ninth, and so on), so we will have done so (and returned our running sum to 0) with a_{99}. a_{100} will then add itself (0) to that sum, and a_{101} will add itself (3) onto that. So we will land at a sum of 3 for s_{101}.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Veritas Prep 10 Year Anniversary Promo Question #3   [#permalink] 19 Sep 2012, 08:01
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