700+ question

[finance]

S is an infinite sequence, in which the value of a particular member of the sequence is equal to the difference between the two preceding members of the sequence. What is the value of the 43rd member of the sequence?

(1) The 1st term is 1.
(2) The 3rd term is 3.

[blaster]

isn't answer D ?

[finance]

No

[newgmattest]

What is official answer please?

[finance]

The OA is A.

I've been wondering if someone has a strong explanation!

[amit2k9]

lets the terms be a1,a2,a3,.....a43.

now a3 = a2- a1

A a1= 1 we have no clue for a2. hence further terms cannot be found out. Not sufficient.

B a3= 3 we have no clue for a1 and a2 as they can be 5,8 or 1,4 and so on. not sufficient.

A+B

a1=1, a2=4, a3=3 a4=-1,a5=-4,a6=-3 and a7=1 a8=4 and repetition.

thus 7*6 = 42 means 42nd term will be = a7 = 1. 43rd term = 4 then.

thus C.

[finance]

amit2k9, your approach seems to be fine, but you still can answer only with the statement 1 given.
If a1= 1, and let a2 be the second term.
a3= a2-1
a4= a2-1-a2= -1
a5= -1-a2+1=-a2
a6= -a2+1
a7= -a2+1+a2=1
a8= 1+a2-1=a2 and it goes on like this.

so for the (6n+1)th term, the value equals 1. Since 42 is 6*7+1, then the 42nd terms will be equal to 1.

[GMATGuruNY]

S is an infinite sequence, in which the value of a particular member of the sequence is equal to the difference between the two preceding members of the sequence. What is the value of the 43rd member of the sequence?

(1) The 1st term is 1.
(2) The 3rd term is 3.

Write out enough of the sequence to determine whether there's a repeating pattern.

Statement 1: 1st term is 1.
Let 2nd term = 2.
The sequence repeats in a pattern of 6: 1,2,1,-1,-2,-1...1,2,1,-1,-2,-1...
Since 6*7=42, the pattern repeats 7 times through term 42.
Since the 43rd term is the beginning of the pattern, the 43rd term is the same as the first term.
Thus, the 43rd term = 1.

Let 2nd term = 50.
The sequence repeats in a pattern of 6: 1,50,49,-1,-50,-49...1,50,49,-1,-50,-49...
Now we can see the nature of the pattern.
Within each pattern of 6, the last 3 terms simply change the signs of the first 3 terms.
Since the difference between the 2nd and 3rd terms is 1, the difference between the 6th and 5th terms must also be 1 -- and this difference yields the 43rd term.
Thus, the 43rd term = 1.
Sufficient.

Statement 2: The 3rd term is 3.
Let 1st term = 1 and 2nd term = 4.
The sequence repeats in a pattern of 6: 1,4,3,-1,-4,-3...1,4,3,-1,-4,-3..
Since 6*7=42, the pattern repeats 7 times through term 42.
Since the 43rd term is the beginning of the pattern, the 43rd term is the same as the first term.
Thus, the 43rd term = 1.

Let 1st term = 2 and 2nd term = 5.
The sequence repeats in a pattern of 6: 2,5,3,-2,-5, -3...2,5,3,-2,-5,-3...
Now we can see the nature of the pattern.
Within each pattern of 6, the last 3 terms simply change the signs of the first 3 terms.
Since the difference between the 2nd and 3rd terms equals the 1st term, the difference between the 6th and 5th terms must also equal the 1st term -- and this difference yields the 43rd term.
Thus, the 43rd term = the 1st term.
Since the 1st term in statement 2 can be any value, insufficient.

The correct answer is A.

[nkumar13]

Let the first no is the sequence be 'a' and second no be 'b'. Thus the sequence will be,
a, b, b-a, -a, b, b+a, a, -b, -b-a, -a, -b, -b+a, a, b, ......

The sequence repeats after 12 numbers, so the 43rd number in the sequence will be
43 = (12x3) + 7 i.e. the seventh digit in the sequence i.e. 'a'.

Statement 1: 1st term is 1
This means 'a' is '1'.
Thus the 43rd number in the sequence is '1'. Hence Sufficient.

Statement 2: 3rd term is 3
This means 'b-a' is '3'.
There is no way to determine a unique value of 'a' just on the basis of the value of 'b-a'.
Hence Insufficient.

Answer is A.

[GMATGuruNY]

Let the first no is the sequence be 'a' and second no be 'b'. Thus the sequence will be,
a, b, b-a, -a, b, b+a, a, -b, -b-a, -a, -b, -b+a, a, b, ......

Careful.
The 5th term = 4th-3rd = -a - (b-a) = -b.
The 6th term = 5th-4th = -b - (-a) = -b+a.
The 7th term = 6th-5th = -b+a - (-b) = a, which is the 1st term.
The 8th term = 7th-6th = a - (-b+a) = b, which is the 2nd term.

Thus, the sequence repeats in a pattern of 6, not 12.

Such errors are easy to make. Luckily, you were still able to determine that statement 1 is sufficient.
Plugging in actual numbers makes such errors less likely.

[cricfan31]

If we assume Tn = T(n-2)- T(n-1)

1. T1 = 1, assume T2 = 2

T3= T1-T2 = -1
T4 = T2-T3 = 2- -1 = 3
Similarly T5 = -4
T5 = 7 and so on

No pattern, shouldnt the answer be E?

[KRISH1985]

Thanks Mitch for the wonderful explanation!:)

[jumsumtak]

Let the first no is the sequence be 'a' and second no be 'b'. Thus the sequence will be,
a, b, b-a, -a, b, b+a, a, -b, -b-a, -a, -b, -b+a, a, b, ......

Careful.
The 5th term = 4th-3rd = -a - (b-a) = -b.
The 6th term = 5th-4th = -b - (-a) = -b+a.
The 7th term = 6th-5th = -b+a - (-b) = a, which is the 1st term.
The 8th term = 7th-6th = a - (-b+a) = b, which is the 2nd term.

Thus, the sequence repeats in a pattern of 6, not 12.

Such errors are easy to make. Luckily, you were still able to determine that statement 1 is sufficient.
Plugging in actual numbers makes such errors less likely.

Mitch,

Yup I agree plugging in numbers is a better trick here..

have 1 more doubt regarding the question.. the question just says the next term is difference of the previous two terms.. it never mentions c=a-b or c=b-a (for a series a,b,c..)

It would be probably safe to assume that the question is talking about absolute difference here i.e. c=|a-b|

though the pattern will repeat the same way, i think we should take the absolute difference here.

comments ?