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

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.


Thanks

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.

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

The first four terms are 3, -3,+3,-3.
sequence for 64 numbers = sum is 0.
Last number will be 3.

Answer= C?

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.

This question would be a little more difficult iw were "2" in options.

how do we get these values ?

B2 = -3
B3 = 2
B4 = -2

hi pushpitkc can you explain pls

Bunuel how can B2 be equal to -B1 = –3 ? can you explain the logic 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.

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!

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

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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