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Let a1, a2, a3....a(n) be a sequence of positive number

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VP
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Let a1, a2, a3....a(n) be a sequence of positive number [#permalink]

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New post 30 Oct 2008, 08:17
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Let a1, a2, a3....a(n) be a sequence of positive number where a1=1 and for n>=1, a(n+1)= a(n)+4. Which of the following is the nth term in the sequence?
a)4n-3
b)4n-1
c)4n
d)4n+1
e)4n+4
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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 08:45
It must be a)4n-3

a(1)=a(0)+4 -> a(0)=-3

a(1)= -3+ 4 * 1 =1
a(2)= -3+ 4 * 2 =5
a(3)= -3+ 4 * 3 =9
a(n)= -3+ 4 * n = 4n-3

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VP
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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 08:52
vishy007 wrote:
It must be a)4n-3

a(1)=a(0)+4 -> a(0)=-3

a(1)= -3+ 4 * 1 =1
a(2)= -3+ 4 * 2 =5
a(3)= -3+ 4 * 3 =9
a(n)= -3+ 4 * n = 4n-3



You are right ... can you explain how you formed the equations ??
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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 08:57
QS says for n>=1, a(n+1)= a(n)+4

Only value we know is a(1), so I checked whether we could derive some info by using only available value.

Thus I arrived at a(1)= a (0+1)= a(0)+4 . Equating this with 1 I got a(0)=-3

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VP
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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 09:03
vishy007 wrote:
QS says for n>=1, a(n+1)= a(n)+4

Only value we know is a(1), so I checked whether we could derive some info by using only available value.

Thus I arrived at a(1)= a (0+1)= a(0)+4 . Equating this with 1 I got a(0)=-3


Thanks vishy ... I am not getting this part
Quote:
a(1)= -3+ 4 * 1 =1
a(2)= -3+ 4 * 2 =5
a(3)= -3+ 4 * 3 =9
a(n)= -3+ 4 * n = 4n-3

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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 09:08
a(2)= -3+ 4 * 2 =5
a(1)= a(0)+ 4 = -3 + 4 * 1
a(2)= a(1) + 4 = a(0) + 4 + 4 = a(0) + 4 * 2 = -3 + 4 * 2
....

Hope this helps.

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VP
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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 09:16
vishy007 wrote:
a(2)= -3+ 4 * 2 =5
a(1)= a(0)+ 4 = -3 + 4 * 1
a(2)= a(1) + 4 = a(0) + 4 + 4 = a(0) + 4 * 2 = -3 + 4 * 2
....

Hope this helps.


Ahh !! Get it now :) Thanks ......
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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 09:29
its simple just plug numbers

n=1 a(2)=5
n=2 a(3)=9
n=3 (a4)=13
n=4 a(5)=17

so when n=4 the series is 17

4n+1

i disagree that a is the answer..

key word in the stem sayd for n>=1 if it had just said n>1 then i agree..

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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 09:35
amitdgr wrote:
Let a1, a2, a3....a(n) be a sequence of positive number where a1=1 and for n>=1, a(n+1)= a(n)+4. Which of the following is the nth term in the sequence?
a)4n-3
b)4n-1
c)4n
d)4n+1
e)4n+4


I found another approach ...

a(n+1)= a(n)+4 implies a(n+1) - a(n) = 4

this means d=4

nth term a(n) = a + (n-1)d = 1 + (n-1)*4 = 1 + 4n - 4 = 4n-3
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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 10:45
i should add, the problem with such type questions is English..i feel like i get confused what they are asking about?

here the question is asking what is a(n)...had i known that it was obvious the ans choice would be 4n-3...

however, what got me was that i was looking for the nth term blindly..not realizing what the question is asking..

if n=4..then the a(5)=4n+1...

so i think it is a good thing to take a few moments just to absorb what the question is asking...

so i guess it will help if you spend the first 30 sec trying to just understand what is being asked..screw the time limit..i think if you are like me the math is mechanical once you have figured what you need to get..

so i guess the biggest challenge for most of us is understanding what is being asked..

i would welcome any tips on how to expiditiously understand what is being asked..

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Re: PS : A Sequence Problem [#permalink]

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New post 30 Oct 2008, 12:12
vishy007 wrote:
It must be a)4n-3

a(1)=a(0)+4 -> a(0)=-3

a(1)= -3+ 4 * 1 =1
a(2)= -3+ 4 * 2 =5
a(3)= -3+ 4 * 3 =9
a(n)= -3+ 4 * n = 4n-3



:cool
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Re: PS : A Sequence Problem   [#permalink] 30 Oct 2008, 12:12
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Let a1, a2, a3....a(n) be a sequence of positive number

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