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In the arithemtic sequence t1, t2, t3, ....tn, ...., t1 = 23

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In the arithemtic sequence t1, t2, t3, ....tn, ...., t1 = 23 [#permalink] New post 26 Feb 2007, 10:55
In the arithemtic sequence t1, t2, t3, ....tn, ...., t1 = 23 and tn = tn-1 - 3 for n>1, what is the value of n when tn=-4?

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 [#permalink] New post 26 Feb 2007, 13:42
10 for me :)

t(n)
= t(n-1) - 3
= t(n-2) - 3*2
= t(n-(n-1)) - 3*(n-1) = t(1) - 3*(n-1) (A)

Then,
t(n) = -4 and t(1) = 23

(A) <=> -4 = 23 - 3*(n-1)
<=> n-1 = 27/3
<=> n = 10
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 [#permalink] New post 26 Feb 2007, 21:54
I think Fig is right, but I'm competely confused.

Fig sensei, can you give me a detailed explanation?
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 [#permalink] New post 26 Feb 2007, 23:32
bz9 wrote:
I think Fig is right, but I'm competely confused.

Fig sensei, can you give me a detailed explanation?


At least, I can try :)

As we know t(1), the first objective is to express t(n) related to the first term of the sequence t(1).

Expression at n-1:
t(n) = t(n-1) - 3

Expression at n-2:
t(n) = t(n-1) - 3
= [t(n-2) - 3] - 3 as t(n-1) = t(n-2) - 3
= t(n-2) - 3*2

Expression at n-3:
t(n) = t(n-2) - 3*2
= [t(n-3) - 3] - 3 as t(n-2) = t(n-3) - 3
= t(n-3) - 3*3

Now, we have identified a pattern. Obersere that, what we remove to have the term n-p : p, this p serve to muplity 3.

Expression at n-p:
t(n) = t(n-p) - 3*(p) (B)

For the next step, we want to have the expressio at 1. Here, we have :
n-p = 1
<=> p = n-1

We replace the value of p in (B):
t(n) = t(n-(n-1)) - 3*(n-1)
= t(1) - 3*(n-1)

Finally, as we know that t(1) = 23 and that t(n) = -4, we plug them in the expression at 1:
-4= 23 - 3*(n-1)
<=> n-1 = 27/3
<=> n = 10 :)
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 [#permalink] New post 27 Feb 2007, 14:14
Thanks Fig,

Perfectly clear now.

In general, does anyone have any material on sequences because that is one of my big weakeness?
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 [#permalink] New post 02 Jun 2007, 14:19
An easier way would have been using the formula:

tn=t1 + (n-1)d where d is the difference between two succeessive terms

Given that tn = tn-1 -3
d=tn-tn-1=-3

t1=23
tn=-4

Therfore, -4=23 + (n-1)-3
3n=23+3+4=30
n=10
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 [#permalink] New post 25 Sep 2007, 05:58
subhen wrote:
An easier way would have been using the formula:

tn=t1 + (n-1)d where d is the difference between two succeessive terms

Given that tn = tn-1 -3
d=tn-tn-1=-3

t1=23
tn=-4

Therfore, -4=23 + (n-1)-3
3n=23+3+4=30
n=10


not sure how u derived the common difference...can you elaborate?
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Re: Sequence [#permalink] New post 25 Aug 2008, 16:00
The question stem says

Tn = Tn-1 - 3 => Tn - Tn-1 = -3 (difference between 2 adjacent terms in AP)

So d = -3

Hope it helps!
Re: Sequence   [#permalink] 25 Aug 2008, 16:00
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In the arithemtic sequence t1, t2, t3, ....tn, ...., t1 = 23

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