sequence.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Thu Sep 23, 2010 7:51 am

interesting.

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Source: Beat The GMAT — Problem Solving | Thu Sep 23, 2010 7:51 am

lokesh r: Master | Next Rank: 500 Posts; Posts: 156; Joined: Sat Sep 04, 2010 2:27 am; Location: Leeds,UK; Thanked: 1 times

by lokesh r » Thu Sep 23, 2010 12:26 pm

It has to be 62

say the sequence question talks about is

k is integer greater than 2
x, kx, k^2x, k^3 x, k^4 x

and k^4 x<1000

For x to be maximum, make k minimum. so k=2, since k cannot be 1 as per question.

5th term = 2^4 x, if x =62

5th term = 988 the largest possible value of 2^4 x less than 1000.

so IMO A.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Thu Sep 23, 2010 12:54 pm

lokesh r wrote:It has to be 62

say the sequence question talks about is

k is integer greater than 2
x, kx, k^2x, k^3 x, k^4 x

and k^4 x<1000

For x to be maximum, make k minimum. so k=2, since k cannot be 1 as per question.

5th term = 2^4 x, if x =62

5th term = 988 the largest possible value of 2^4 x less than 1000.

so IMO A.

No man the answer is 63
I think they are are considering 0
As zero is a positive integer.
I don't know i am right or wrong but i only find this reason for the answer to be 63.

I m editing this post just to write down that i put in 5 minutes and 6 seconds to get the right answer and still got it wrong.

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lokesh r: Master | Next Rank: 500 Posts; Posts: 156; Joined: Sat Sep 04, 2010 2:27 am; Location: Leeds,UK; Thanked: 1 times

by lokesh r » Thu Sep 23, 2010 1:02 pm

goyalsau wrote:
lokesh r wrote:It has to be 62

say the sequence question talks about is

k is integer greater than 2
x, kx, k^2x, k^3 x, k^4 x

and k^4 x<1000

For x to be maximum, make k minimum. so k=2, since k cannot be 1 as per question.

5th term = 2^4 x, if x =62

5th term = 988 the largest possible value of 2^4 x less than 1000.

so IMO A.
No man the answer is 63
I think they are are considering 0
As zero is a positive integer.
I don't know i am right or wrong but i only find this reason for the answer to be 63.

I m editing this post just to write down that i put in 5 minutes and 6 seconds to get the right answer and still got it wrong.

yes you are right..

zero is neither +ve nor -ve.

so it is ( zero , 1,2,3....62) , that is 63 terms..

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Thu Sep 23, 2010 1:06 pm

lokesh r wrote:
goyalsau wrote:
lokesh r wrote:It has to be 62

say the sequence question talks about is

k is integer greater than 2
x, kx, k^2x, k^3 x, k^4 x

and k^4 x<1000

For x to be maximum, make k minimum. so k=2, since k cannot be 1 as per question.

5th term = 2^4 x, if x =62

5th term = 988 the largest possible value of 2^4 x less than 1000.

so IMO A.
No man the answer is 63
I think they are are considering 0
As zero is a positive integer.
I don't know i am right or wrong but i only find this reason for the answer to be 63.

I m editing this post just to write down that i put in 5 minutes and 6 seconds to get the right answer and still got it wrong.
yes you are right..

zero is neither +ve nor -ve.

so it is ( zero , 1,2,3....62) , that is 63 terms..

you are saying Zero is neither +ve nor -ve
but as i know Zero is +ve.

Please correct me if i am wrong.

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lokesh r: Master | Next Rank: 500 Posts; Posts: 156; Joined: Sat Sep 04, 2010 2:27 am; Location: Leeds,UK; Thanked: 1 times

by lokesh r » Thu Sep 23, 2010 1:23 pm

goyalsau wrote:
lokesh r wrote:
goyalsau wrote:
lokesh r wrote:It has to be 62

say the sequence question talks about is

k is integer greater than 2
x, kx, k^2x, k^3 x, k^4 x

and k^4 x<1000

For x to be maximum, make k minimum. so k=2, since k cannot be 1 as per question.

5th term = 2^4 x, if x =62

5th term = 988 the largest possible value of 2^4 x less than 1000.

so IMO A.
No man the answer is 63
I think they are are considering 0
As zero is a positive integer.
I don't know i am right or wrong but i only find this reason for the answer to be 63.

I m editing this post just to write down that i put in 5 minutes and 6 seconds to get the right answer and still got it wrong.
yes you are right..

zero is neither +ve nor -ve.

so it is ( zero , 1,2,3....62) , that is 63 terms..
you are saying Zero is neither +ve nor -ve
but as i know Zero is +ve.

Please correct me if i am wrong.

Please check this below link.

https://answers.yahoo.com/question/index ... 157AAyW3ZI

would be happy if any expert can comment on this.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Thu Sep 23, 2010 11:55 pm

Well Lokesh I am still not clear With Zero
Because problem says all those numbers which are not negative
as per the definition of ZERO
IT is not -ve
But if it is not even positive Then how can we consider zero in the above Case then the answer must be 62.

completely Confused

Experts please comment on this..

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