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In the sequence S of numbers, each term after the first two terms is

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In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 26 Feb 2014, 02:21
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In the sequence S of numbers, each term after the first two terms is the sum of the two immediately preceding terms. What is the 5th term of S?

(1) The 6th term of S minus the 4th term equals 5.
(2) The 6th term of S plus the 7th term equals 21.

Data Sufficiency
Question: 111
Category: Arithmetic Sequences
Page: 161
Difficulty: 650


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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 26 Feb 2014, 02:22
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SOLUTION

In the sequence S of numbers, each term after the first two terms is the sum of the two immediately preceding terms. What is the 5th term of S?

\(s_n=s_{n-1}+s_{n-2}\), for \(n>2\).

(1) The 6th term of S minus the 4th term equals 5 --> \(s_6-s_4=5\) --> \((s_5+s_4)-s_4=5\) --> \(s_5=5\). Sufficient.

(2) The 6th term of S plus the 7th term equals 21 --> \(s_6+s_7=21\) --> \(s_6+(s_6+s_5)=21\). Since we don't know \(s_6\) we cannot find \(s_5\). Not sufficient.

Answer: A.
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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 27 Feb 2014, 22:17
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Lets assume the first two terms to be x and y
Then rest of the numbers in the series will be x, y, x+y, x+2y, 2x+3y, 3x+5y, 5x+8y, ...

A.) (3x+5y) - (x+2y) = 5
2x+3y = 5 which is indeed the fifth term

SUFFICIENT

B.) (3x+5y) +(5x+8y) = 21
8x+13y = 21
its the 8th term

INSUFFICIENT

Answer - A
Difficulty - 650
Time taken - 1:59
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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 27 Feb 2014, 22:47
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b2bt wrote:
Lets assume the first two terms to be x and y
Then rest of the numbers in the series will be x, y, x+y, x+2y, 2x+3y, 3x+5y, 5x+8y, ...

A.) (3x+5y) - (x+2y) = 5
2x+3y = 5 which is indeed the fifth term

SUFFICIENT

B.) (3x+5y) +(5x+8y) = 21
8x+13y = 21
its the 8th term

INSUFFICIENT

Answer - A
Difficulty - 650
Time taken - 1:59


Or you can simply say that the terms are t1, t2, ....

(1) The 6th term of S minus the 4th term equals 5.
t6 - t4 = 5
But t6 = t4 + t5
t6 - t4 = t5 = 5
Sufficient

(2) The 6th term of S plus the 7th term equals 21.
t6 + t7 = 21
t6 + t6 + t5 = 21
We don't know t6 and hence we cannot find t5.
Not sufficient

Answer (A)
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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 19 Dec 2015, 05:33
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


In the sequence S of numbers, each term after the first two terms is the sum of the two immediately preceding terms. What is the 5th term of S?

(1) The 6th term of S minus the 4th term equals 5.
(2) The 6th term of S plus the 7th term equals 21.

When you modify the original condition and the question, they become S_n=S_(n-1)+S_(n-2). When you substitute n=5, or S_5=S_4+S_3 or n=6, S_6=S_5+S_4 becomes S_5=S_6-S_4. (The reason why you substitute n=5,6 is S_5=?) Therefore, in 1), S_5-S_4=5 and this is S_5=5, which is sufficient. Therefore, the answer is A.


-> Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 23 May 2018, 19:47
Bunuel wrote:
SOLUTION

In the sequence S of numbers, each term after the first two terms is the sum of the two immediately preceding terms. What is the 5th term of S?

\(s_n=s_{n-1}+s_{n-2}\), for \(n>2\).

(1) The 6th term of S minus the 4th term equals 5 --> \(s_6-s_4=5\) --> \((s_5+s_4)-s_4=5\) --> \(s_5=5\). Sufficient.

(2) The 6th term of S plus the 7th term equals 21 --> \(s_6+s_7=21\) --> \(s_6+(s_6+s_5)=21\). Since we don't know \(s_6\) we cannot find \(s_5\). Not sufficient.

Answer: A.



Can anyone please explain how S7 = S6+S5 as i did not get it?

Instead of x, y, x+y, x+2y, 2x+3y, 3x+5y, 5x+8y, ..., Can i consider x,y, x+y, 2x+y, 2x+2y, 4x+3y,4x+4y?
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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 23 May 2018, 22:01
Raxit85 wrote:
Bunuel wrote:
SOLUTION

In the sequence S of numbers, each term after the first two terms is the sum of the two immediately preceding terms. What is the 5th term of S?

\(s_n=s_{n-1}+s_{n-2}\), for \(n>2\).

(1) The 6th term of S minus the 4th term equals 5 --> \(s_6-s_4=5\) --> \((s_5+s_4)-s_4=5\) --> \(s_5=5\). Sufficient.

(2) The 6th term of S plus the 7th term equals 21 --> \(s_6+s_7=21\) --> \(s_6+(s_6+s_5)=21\). Since we don't know \(s_6\) we cannot find \(s_5\). Not sufficient.

Answer: A.



Can anyone please explain how S7 = S6+S5 as i did not get it?

Instead of x, y, x+y, x+2y, 2x+3y, 3x+5y, 5x+8y, ..., Can i consider x,y, x+y, 2x+y, 2x+2y, 4x+3y,4x+4y?


The stem says that each term after the first two terms is the sum of the two immediately preceding terms. So, the seventh term \(s_7\) equals to the sum of the two immediately preceding terms, which are \(s_5\) and \(s_6\): \(s_7=s_6+s_5\).

If you consider the first term to be x and the second term to be y, then:
1st term = x;
2nd term = y;
3rd term = x + y;
4th term = y + (x + y) = 2y + x;
5th term = (x + y) + (2y + x) = 3y + 2x;
...
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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 19 Jan 2019, 01:21
Bunuel wrote:

The stem says that each term after the first two terms is the sum of the two immediately preceding terms. So, the seventh term \(s_7\) equals to the sum of the two immediately preceding terms, which are \(s_5\) and \(s_6\): \(s_7=s_6+s_5\).

If you consider the first term to be x and the second term to be y, then:
1st term = x;
2nd term = y;
3rd term = x + y;
4th term = y + (x + y) = 2y + x;
5th term = (x + y) + (2y + x) = 3y + 2x;
...


okay so i did this
6th term = 3x + 5y
7th term = 5x + 8y

St1 -- 6th - 4th = 5
3x + 5y - x - 2y => 2x + 3y = 5 --looks like x n y should be 1

St2 -- 6th + 7th = 21
3x + 5y +v5x + 8y => 8x + 13y = 21 --looks like x n y should be 1

could you please let me know why above is wrong ?
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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 19 Jan 2019, 01:39
gmatns2018 wrote:
Bunuel wrote:

The stem says that each term after the first two terms is the sum of the two immediately preceding terms. So, the seventh term \(s_7\) equals to the sum of the two immediately preceding terms, which are \(s_5\) and \(s_6\): \(s_7=s_6+s_5\).

If you consider the first term to be x and the second term to be y, then:
1st term = x;
2nd term = y;
3rd term = x + y;
4th term = y + (x + y) = 2y + x;
5th term = (x + y) + (2y + x) = 3y + 2x;
...


okay so i did this
6th term = 3x + 5y
7th term = 5x + 8y

St1 -- 6th - 4th = 5
3x + 5y - x - 2y => 2x + 3y = 5 --looks like x n y should be 1

St2 -- 6th + 7th = 21
3x + 5y +v5x + 8y => 8x + 13y = 21 --looks like x n y should be 1

could you please let me know why above is wrong ?


2x + 3y = 5 has INFINITELY MANY solutions. It's an equation with TWO unknowns. For example, x = 0 and y =5/3.
8x + 13y = 21 has INFINITELY MANY solutions. It's an equation with TWO unknowns. For example, x = 0 and y =21/3.
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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 26 Feb 2019, 07:08
Hi,

I am not able to understand why is not d).

When I first read the problem, the Fibonacci sequence instantly came to my mind: 0,1,2,3,5,8,13,21...
The only way I found to start the sequence having the 6th and 7th terms summing up 21 is starting by 0. Therefore I know that 5 is my solution as it is the 5th term.

That (probably incorrect) logic also validates the first statement.

Where am I making the mistake?

Please Bunuel help me! I want to be like you!
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Re: In the sequence S of numbers, each term after the first two terms is  [#permalink]

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New post 26 Feb 2019, 07:13
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BunuelWannabe wrote:
Hi,

I am not able to understand why is not d).

When I first read the problem, the Fibonacci sequence instantly came to my mind: 0,1,2,3,5,8,13,21...
The only way I found to start the sequence having the 6th and 7th terms summing up 21 is starting by 0. Therefore I know that 5 is my solution as it is the 5th term.

That (probably incorrect) logic also validates the first statement.

Where am I making the mistake?

Please Bunuel help me! I want to be like you!


That's not the only sequence satisfying the stem and the second statement. For example, check 1, 1, 2, 3, 5, 8, 13, ...
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Re: In the sequence S of numbers, each term after the first two terms is   [#permalink] 26 Feb 2019, 07:13
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