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Problem solving - 600 level question

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Problem solving - 600 level question [#permalink] New post 23 May 2010, 08:24
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An Arithmetic sequence is defined as a sequence in which each term after the first is equal to the sum of the preceding term and a constant. If the list of letters shown is an arithmetic sequence, which of the following must also be an arithmetic sequence?

I a-1, b-2, c-3, d-4, e-5
II 3a, 3b, 3c, 3d, 3e
III a2, b2 ,c2, d2, e2
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Re: Problem solving - 600 level question [#permalink] New post 23 May 2010, 09:11
shekar123 wrote:
An Arithmetic sequence is defined as a sequence in which each term after the first is equal to the sum of the preceding term and a constant. If the list of letters shown is an arithmetic sequence, which of the following must also be an arithmetic sequence?

I a-1, b-2, c-3, d-4, e-5
II 3a, 3b, 3c, 3d, 3e
III a2, b2 ,c2, d2, e2



Are you missing something here ?

I think you missed the part that a,b,c,d,e is an arithmetic sequence. Or something about a,b,c,d,e ? Without that information it would be impossible to answer this question
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Re: Problem solving - 600 level question [#permalink] New post 23 May 2010, 14:49
dimitri92 wrote:
shekar123 wrote:
An Arithmetic sequence is defined as a sequence in which each term after the first is equal to the sum of the preceding term and a constant. If the list of letters shown is an arithmetic sequence, which of the following must also be an arithmetic sequence?

I a-1, b-2, c-3, d-4, e-5
II 3a, 3b, 3c, 3d, 3e
III a2, b2 ,c2, d2, e2



Are you missing something here ?

I think you missed the part that a,b,c,d,e is an arithmetic sequence. Or something about a,b,c,d,e ? Without that information it would be impossible to answer this question



I did not miss anything here
OA is I and II
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Re: Problem solving - 600 level question [#permalink] New post 23 May 2010, 22:20
shekar123 wrote:
dimitri92 wrote:
shekar123 wrote:
An Arithmetic sequence is defined as a sequence in which each term after the first is equal to the sum of the preceding term and a constant. If the list of letters shown is an arithmetic sequence, which of the following must also be an arithmetic sequence?

I a-1, b-2, c-3, d-4, e-5
II 3a, 3b, 3c, 3d, 3e
III a2, b2 ,c2, d2, e2



Are you missing something here ?

I think you missed the part that a,b,c,d,e is an arithmetic sequence. Or something about a,b,c,d,e ? Without that information it would be impossible to answer this question



I did not miss anything here
OA is I and II

Whats the source of this question?
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Re: Problem solving - 600 level question [#permalink] New post 23 May 2010, 22:55
dimitri92 wrote:
shekar123 wrote:
An Arithmetic sequence is defined as a sequence in which each term after the first is equal to the sum of the preceding term and a constant. If the list of letters shown is an arithmetic sequence, which of the following must also be an arithmetic sequence?

I a-1, b-2, c-3, d-4, e-5
II 3a, 3b, 3c, 3d, 3e
III a2, b2 ,c2, d2, e2



Are you missing something here ?

I think you missed the part that a,b,c,d,e is an arithmetic sequence. Or something about a,b,c,d,e ? Without that information it would be impossible to answer this question


I have just highlighted the part which is missing. I think what you forgot to provide was that a,b,c,d,e was the list of arithmetic sequence shown below the question. This is crucial piece of information because if we know that a.b,c,d,e is an arithmetic sequence, we can deduce conclusions about I, II and III. I hope I am getting my point across.
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Re: Problem solving - 600 level question [#permalink] New post 24 May 2010, 05:18
I didnt got the answer correct.

For example let constant be one and first term be 1 then the series will be
1,2,3,4,5,6,7,8....

With statement I the series will become,

1,1,1,1,1,1,1,1..... which is NOT an arithematic series defined in the question.

With statement II the series will become,

3,6,9,12,.... which is an arithematic series defined in the question.

With statement III the series will become,

2,4,6,8,... which is again an arithematic series....

So, the answer should be..

II and III

Please correct me if i am wrong.
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Re: Problem solving - 600 level question [#permalink] New post 24 May 2010, 06:39
I just want to make sure I understand the method in which you came to the sequences so I'm double checking one calcuation I coulnd't qualify on my own, Statement I.

I'm not 100% here but using the sequence you've identified; 1, 2, 3, 4, 5...
Shouldn't that make statement I. = 0, 0, 0, 0, 0? instead of 1, 1, 1, 1, 1?

I could be wrong but this is what I'm understanding;

Statement I. a-1, b-2, c-3, d-4, e-5
So given that in our sequence; a = 1, b = 2, c = 3, d = 4, e = 5 then that would provide the following;

(1)-1, (2)-2, (3)-3, (4)-4, (5)-5 = 0, 0, 0, 0, 0

With this understanding then if statement I. was listed as the following it would yeild 1, 1, 1, 1, 1
b-1, c-1, d-1, e-1, f-1
Which would give us:
(2)-1, (3)-2, (4)-3, (5)-4, (6)-5 = 1, 1, 1, 1, 1

Is this correct or am I not getting the concept?
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Re: Problem solving - 600 level question [#permalink] New post 25 May 2010, 04:32
yes you are correct... but our problem doesnt get resolved with this...
Let it be 0,0,0,0,0,0,...
Still its not arithmetic series....
So statement II and Statement III must be the correct answer.....
Am i wrong or I am making some mistake in understanding???
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Re: Problem solving - 600 level question [#permalink] New post 25 May 2010, 09:58
amitjash wrote:
yes you are correct... but our problem doesnt get resolved with this...
Let it be 0,0,0,0,0,0,...
Still its not arithmetic series....
So statement II and Statement III must be the correct answer.....
Am i wrong or I am making some mistake in understanding???


Hi,
Lets define A,B,C,D,E... as AP and d as the value that is added to get the next term
so that B = A+d and C = B+d
Then below is still AP (arithmetic progression).

0000000
1111111

In this case d is 0 and the next term is still the sum of previous term and constant(0).

Now, for AP
B = A+d . EQ 1
C = B+d . EQ 2
therefore, d = C-B

if you replace d in EQ 1
B = A + C - B
or 2B = A+C

In other word, twice of second term = sum of first and third
Lets see if this holds true for I
2(B-2) = A-1 + C - 3
2B - 4 = A + C -4
2B = A + C

So first is correct

For option II

2 (3B) = 3A + 3C
2B = A + C
II is also correct.

Option III is square of the each term in AP
suppose if AP is 1,2,3,4,5 then it would be
1,4,9,16,25 which is not an AP

Therefore only I and II are correct.

rgds Anirudh.
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Re: Problem solving - 600 level question [#permalink] New post 30 May 2010, 12:15
amitjash yes you are correct II and III are the answers, I just wanted to make sure I was "testing" the sequence correctly, when you stated 1, 1, 1, 1, 1 when I got 0, 0, 0, 0, 0 when I did the problem myself I wanted to make sure I wasn't incorrectly executing.

Thanks for all your replies and explanations!
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Re: Problem solving - 600 level question [#permalink] New post 30 May 2010, 22:09
An Arithmetic sequence is defined as a sequence in which each term after the first is equal to the sum of the preceding term and a constant. If the list of letters shown is an arithmetic sequence, which of the following must also be an arithmetic sequence?

I a-1, b-2, c-3, d-4, e-5
II 3a, 3b, 3c, 3d, 3e
III a2, b2 ,c2, d2, e2

a,b,c,d,e are in AP.

Consider k=b-a=c-b=d-c=e-d = difference between successive numbers.

For I.
b-2 - (a-1) = b-a -1 = k-1
c-3 - (b-2) = c-b -1 = k-1
Similarly for others
So difference between two successive numbers is k-1, always same. Like original sequence, this one is in AP.

For II.

3b-3a = 3k
3c-3b = 3k

So difference between two successive numbers is 3k, always same.
Like original sequence, this one is in AP.

For III,

b2 - a2 = (b+a)(b-a) = (b+a)k .
So difference between two successive numbers is not always same.
This one is not in AP.
Re: Problem solving - 600 level question   [#permalink] 30 May 2010, 22:09
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