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Math Expert V
Joined: 02 Sep 2009
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Veritas Prep 10 Year Anniversary Promo Question #3  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 66% (02:33) correct 34% (02:59) wrong based on 221 sessions

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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

To participate, please make sure you provide the correct answer (A,B,C,D,E) and explanation that clearly shows how you arrived at it.
Winners will be announced the following day at 10 AM Pacific/1 PM Eastern Time.

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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Posts: 62439
Re: Veritas Prep 10 Year Anniversary Promo Question #3  [#permalink]

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1
3

Winner:

yogeshwar007

Official Explanation:

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]

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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]

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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

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Re: Veritas Prep 10 Year Anniversary Promo Question #3  [#permalink]

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1
Ans is C,

This would be a series of -3 , 0, 3,0 ... .

Originally posted by Vips0000 on 18 Sep 2012, 09:10.
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]

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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]

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1

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

Originally posted by jirayr on 18 Sep 2012, 09:14.
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]

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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]

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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.

Originally posted by chetanachethu on 18 Sep 2012, 09:47.
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]

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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.

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Re: Veritas Prep 10 Year Anniversary Promo Question #3  [#permalink]

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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.

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Re: Veritas Prep 10 Year Anniversary Promo Question #3  [#permalink]

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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

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Re: Veritas Prep 10 Year Anniversary Promo Question #3  [#permalink]

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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 [/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]

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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]

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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 $$s_{101}$$?[/b]

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

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

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Re: Veritas Prep 10 Year Anniversary Promo Question #3  [#permalink]

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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]

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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]

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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]

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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]

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We have a winner!

Winner:

yogeshwar007

Official Explanation:

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 $$s_{101}$$.
_________________ Re: Veritas Prep 10 Year Anniversary Promo Question #3   [#permalink] 19 Sep 2012, 08:01

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