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For all n such that n is a positive integer, the terms of a certain

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New post 08 Jun 2015, 08:08
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For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.

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Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 08 Jun 2015, 08:39
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Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.


B1 = 3
B2 = -3
B3 = 2
B4 = -2
B5 = 3 and from this moment we see that pattern will be repeat each four numbers
65 / 4 = 4*16 + 1

so B65 will be equal to 3 (16 repetions + 1 more calculation)

Answer is C
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Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 08 Jun 2015, 08:55
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The correct answer= C.

The first four terms in the sequence are 3, -3,+3,-3. Repeat teh sequence for 16 times. You get 64 numbers and the sum is 0. The last number will be 3.

KR.

Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.

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Kudos to you, for helping me with some KUDOS.

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Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 08 Jun 2015, 09:10
1
First four terms are repeating B1=3, B2=-3, B3=2, B4=-2....

There are 16 such repetitions equal to 64 and the 65th term is B65=B1=3.


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Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 09 Jun 2015, 01:00
1
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8


Ans: C

Solution: by given condition we start from B1=3, B2=-3, B3=2, B4=-2, B5=3, B6=-3 B7=2, B8=-2....... and so on
so this sequence repeat first 4 terms with the cycle of 4 terms. and there sum is 0.
so now first 65 terms had 16 full cycle and one extra term B65=3

so the sum of first 64 terms is 3.
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Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 10 Jun 2015, 00:56
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Bn=Bn−1+5 if n is odd and greater than 1;
Bn=−Bn−1 if n is even;
B1=3
\(B_1 =3\)
\(B_2 =-B_1=-3\) because 2 is even
\(B_3 =B_2+5=2\)
\(B_4 =-B_3=-2\)
\(B_5 =B_4+5=3\)
\(B_6 =-B_5=-3\) etc.
so it is \(3-3+2-2+3-3+...\) 35 times, with every two numbers summing to 0. For the first 65 terms, there are 32 pairs and then the 65th term. The 65th term is 3, because it starts the 33th pair, and the start of every odd pair is 3. So the answer is C
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Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 14 Jun 2015, 03:47
1
The first four terms are 3, -3,+3,-3.
sequence for 64 numbers = sum is 0.
Last number will be 3.

Answer= C?
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Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 15 Jun 2015, 02:45
1
1
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Let's list the first few terms of the sequence according to the given rules:
B1 = 3
B2 = -B1 = –3
B3 = B2 + 5 = –3 + 5 = 2
B4 = -B3 = –2
B5 = B4 + 5 = –2 + 5 = 3
…etc.

Note that the pattern is a four-term repeat: 3, –3 , 2, –2 . Also note that the sum of this repeating group is (3) + (–3) + (2) + (–2) = 0. This repeating group will occur 16 times through term number 64. Thus, the sum of the first 64 terms will be 0. This leaves the 65th term, which will have the same value as B1 : 3. Therefore, the sum of the first 65 terms is 3.

The correct answer is C.
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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

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Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 08 Sep 2018, 01:42
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.


This question would be a little more difficult iw were "2" in options.
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For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 08 Sep 2018, 05:12
Harley1980 wrote:
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.


B1 = 3
B2 = -3
B3 = 2
B4 = -2
B5 = 3 and from this moment we see that pattern will be repeat each four numbers
65 / 4 = 4*16 + 1

so B65 will be equal to 3 (16 repetions + 1 more calculation)

Answer is C


how do we get these values ?

B2 = -3
B3 = 2
B4 = -2

hi pushpitkc can you explain pls
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Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 08 Sep 2018, 08:07
Bunuel wrote:
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Let's list the first few terms of the sequence according to the given rules:
B1 = 3
B2 = -B1 = –3
B3 = B2 + 5 = –3 + 5 = 2
B4 = -B3 = –2
B5 = B4 + 5 = –2 + 5 = 3
…etc.

Note that the pattern is a four-term repeat: 3, –3 , 2, –2 . Also note that the sum of this repeating group is (3) + (–3) + (2) + (–2) = 0. This repeating group will occur 16 times through term number 64. Thus, the sum of the first 64 terms will be 0. This leaves the 65th term, which will have the same value as B1 : 3. Therefore, the sum of the first 65 terms is 3.

The correct answer is C.



Bunuel how can B2 be equal to -B1 = –3 ? :? can you explain the logic pls :)
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For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 08 Sep 2018, 18:46
1
dave13 wrote:
Harley1980 wrote:
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8


B1 = 3
B2 = -3
B3 = 2
B4 = -2
B5 = 3 and from this moment we see that pattern will be repeat each four numbers
65 / 4 = 4*16 + 1

so B65 will be equal to 3 (16 repetions + 1 more calculation)

Answer is C

how do we get these values ?

B2 = -3
B3 = 2
B4 = -2

hi pushpitkc can you explain pls

dave13 , you will have to deal with my pinch-hitting for pushpitkc :)

I can't quite tell what part does not make sense, so I may give you too much info. The way that the instructions are written may be throwing you off.

This problem is like a recipe for a sequence. The "recipe" tells us how to calculate each term.

The "ingredients" are terms with even- and odd-numbered subscripts. Even- and odd-numbered terms have different rules.

We specify each term by using a subscript, \(B_{n}\). The first term is \(B_1\), where \(n=1\). Second term is \(B_2\), where \(n=2\)

As mentioned, if the subscript is an odd number, we follow one rule. If subscript is even, we follow another rule.

Rule (1): IF \(n\) in \(B_{n}\) is ODD and greater than 1, use this rule: \(B_{n}=B_{(n-1)}+5\)

That rule means: to calculate the value of THIS term, use the value of the preceding term and add 5 to it.
\((n-1)\) means "use the term with a subscript that is 1 less than this term's subscript"

If we were calculating \(B_5\), we would plug in the value of \(B_4\).(then add 5) \(B_5\):

\(B_{(n-1)}=B_{(5-1)}=B_4\)
\(B_5\) thus will equal \(B_4+5\)

Rule (2) IF \(n\) in \(B_{n}\) is EVEN, use this rule: \(B_{n} = -(B_{(n-1)})\)
Meaning: take the value of the preceding term (n-1), and negate it.

1) Start calculating sequence terms. We need a pattern. Given:
\(B_1=3\)
Next term, \(B_2\)? Use rule for evens. Negate the preceding term \(B_2=-(B_{(n-1)})=-(B_{(2-1)})=-(B_1)=-(3)\)
That is, \(B_2=-3\)
\(B_3\)? Use the rule for odds

\(B_1=3\)
\(B_2=-(B_{(2-1)})=-(B_1)=-3\)
\(B_3=(B_{(n-1)}+5)=(B_2+5)=(-3+5)=2\)
\(B_4=-(B_3)=-(2)=-2\)
\(B_5=(B_4+5)=(-2+5)=3\)
\(B_6=-(B_5)=-(3)=-3\)
\(B_7=(B_6+5)=(-3+5)=2\)
\(B_8=-(B_7)=-(2)=-2\)

We have a pattern that repeats in cycles of 4 terms: \(3, -3, 2, -2\)

2) Cyclicity
We need the sum of "the first 65 terms." We will use the cycles of four terms to find that sum. Cyclicity in this case is like integer powers and units digits cyclicity.

We need to know where, in the cycle of four terms, \(B_{65}\) will "fall."

To find where \(B_{65}\) falls in the cycle of four terms: Divide the subscript "65" by the cyclicity of "4"
Remainder? Match the remainder to a subscript. The term whose subscript equals the remainder has the same value as \(B_{65}\)

No remainder? Then \(B_{65}\) has the same value as \(B_4\) (the term is a multiple of 4)

\(\frac{65}{4}=16\) + remainder 1
Thus \(B_{65}=B_1=3\)

3) SUM??
If we were literally to sum all the terms, the first 8 would be
\(3+(-3)+2+(-2)+(3)+(-3)+2+(-2)=\)
No thanks. We do not need to sum all the terms.
Every 4 terms, the sum of terms = 0: \(3+(-3)+2+(-2)=0\)

\(B_{65}\) had a "remainder 1." One term before it, \(B_{64}\) was the last term in a cycle of four numbers

So up to the 64\(^{th}\) term, from \(B_1\) to \(B_{64}\), the sum is 0. (16 cycles of four terms; every 4 terms= 0).
\(B_{65}=3\)

\((0+0+0+0+0+0. . . +3)=3\)
The sum of the first 65 terms \(= 3\)
Hope that helps.:-)
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For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 25 Jan 2019, 17:54
Hello!

Does anyone know where can I find more problems like this?

\(B_n = B_{n-1} + 5\)
\(Bn = -B_{n-1}\)


It has been very complicated for me to interpret the formulas, I take a lot of time to understand because I get confused with the letter terms.

Any advice on how to improve this?

Kind regards!
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For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 25 Jan 2019, 19:13
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.


In these types of questions, the condition is given, if we can find a pattern we should be good, \(B_1 = 3\)

when n is odd, then \(B_n = B_{n-1} + 5\)
when n is even, then \(Bn = -B_{n-1}\)

So series will be
n = 1,3,5,7,9

n1 = 3 , n3 = 2, n5 = 3, n7 = 2

n = 2,4,6,8,10

n2 =-3, n4= -2, n6 = -3, n8 = -2

So now we found the pattern, 3,-3,2,-2

Now the 65th term will be odd, which will be \(n_{64}\) +5

Answer is 3
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For all n such that n is a positive integer, the terms of a certain  [#permalink]

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New post 25 Jan 2019, 19:21
1
jfranciscocuencag wrote:
Hello!

Does anyone know where can I find more problems like this?

\(B_n = B_{n-1} + 5\)
\(Bn = -B_{n-1}\)


It has been very complicated for me to interpret the formulas, I take a lot of time to understand because I get confused with the letter terms.

Any advice on how to improve this?

Kind regards!


Hi,

Would like to share my few cents on this.

When you are practicing for questions on Sequences, always try to note down what is been asked and then proceed accordingly,
You might take time initially but in the end it will pay well.

Make a list of the particular elements and then proceed.
Dont be in a rush to solve such questions, because one incorrect step can cause a trouble.

You can practice the questions from this list
https://gmatclub.com/forum/search.php?s ... tag_id=112

Try with the OG questions first, then you can go the next level of questions.

Let me know if this helps you.
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

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For all n such that n is a positive integer, the terms of a certain   [#permalink] 25 Jan 2019, 19:21
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