A sequence is given by the rule an = |a_(n-2)| - |a(n-1)|

Question

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 Veri tas 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: https://gmatclub.com/forum/veritas-prep ... 38806.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.

Bunuel · Accepted Answer

Winner:    yogeshwar007

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} .

fameatop · Answer

Answer is E as the sum of any first two terms is 3 & then 6, 6, 12, 12 & so on

yogeshwar007 · Answer

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)

Vips0000 · Answer

Ans is C,

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

Aspirant18 · Answer

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

jirayr · Answer

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

catfreak · Answer

IMo - C

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

sum = 3

chetanachethu · Answer

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.

bagdbmba · Answer

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   gibes a quotient 99 and thus sum of a1+a2+..a99=0  and a100+a101=|0|+|3|=3

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

So Answer is 3.

neha087 · Answer

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

dexerash · Answer

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.

nadeemkamal · Answer

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    leaves a remainder 2; therefore sum (a1+a2+a3+...a99)=0
S101=0+0+3
=3, (c)

Vish24 · Answer

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

nnitingarg · Answer

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

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?

chris558 · Answer

Answer is C!

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

101/3=99 R2

0+3=3

huyentran · Answer

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

Aponcci · Answer

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

jsahni123 · Answer

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

Bunuel · Answer

We have a winner!

Winner:    yogeshwar007

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} .

A sequence is given by the rule an = |a_(n-2)| - |a(n-1)|

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A sequence is given by the rule an = |a_(n-2)| - |a(n-1)|

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards