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# In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn

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In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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20 Mar 2012, 09:00
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In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn= tn-1 - 3 for each n > 1. What is the value of n when tn = -4?

A. -1
B. 7
C. 10
D. 14
E. 20
[Reveal] Spoiler: OA

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20 Mar 2012, 09:42
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well.....t2 =t1-3 =23-3=20

t3=t2-3=20-3=17

So every time we n increases tn decreases by 3.

Since t1=23 we have 23-3=20-3=17-3=14-3=11-3=8-3=5-3=2-3=-1-3=-4!VOILA

So n=10 ten times substructing 3 from 23 to reach -4.

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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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20 Mar 2012, 13:44
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enigma123 wrote:
In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn= tn-1 - 3 for each n > 1. What is the value of n when tn = -4?

A. -1
B. 7
C. 10
D. 14
E. 20

$$t_n=t_{n-1}-3$$ means that each term is 3 less than the previous term. Now, the difference between $$t_1=23$$ and $$t_n=-4$$ is $$23-(-4)=27$$, so we moved $$\frac{27}{3}=9$$ terms from $$t_1$$, so from $$t_1$$ to $$t_{10}$$.

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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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12 Apr 2012, 10:24
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tn= tn-1 - 3 means that d=-3

tn=t1 +d(n-1)
-4=23-3(n-1)
-30=-3n
n=10
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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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22 Apr 2014, 02:15
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23... 20.... 17... 14... 11.... 8... 5... 2.... -1.... -4

-4 is the 10th term

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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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03 Feb 2015, 01:44
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I also did the same as Paresh, but like this:

We know that t1 = 23
So, using the given formula we have:
t1=(t1-1) -3 =23
t0 - 3 = 23
t0= 26

The sam way we find that t2= 20

It seems that the sequence goes like this:
t0 = 26
t1 = 23
t2 = 20
t3 = 17
t4 = 14
t5 = 11
t6 = 8
t7 = 5
t8 = 2
t9 = -1
t10 = -4

So, our ANS is C.

However, I did do it wrong because, by the way I saw it writen, I though that the whole formula equaled to 23 (t1, t2, t3, ..., tn, t1=23). I didn't see any relationship as to why this is 23 (it is an addition, is it a multiplication?). So, then I thought that what was meant is that this sequence is doing a circle, going from t1 to t1 again, and there are 23 numbers in the sequence. So, I though we needed to find tn.

Hopefully the gmat would show in a more clear way that t1=23 and not the whole sequence..
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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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09 Jan 2017, 00:59
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t1=23
t2=t1-3=20
t3=t2-3=17 and so on...
Here is when we need to consider the formula for AP as we know the common difference is -3

tn=t1 + d(n-1)

given, tn=-4
-4=23 + (-3) (n-10) >> n=10

Ans : C
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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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09 Jan 2017, 03:17
enigma123 wrote:
In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn= tn-1 - 3 for each n > 1. What is the value of n when tn = -4?

A. -1
B. 7
C. 10
D. 14
E. 20

Its a normal arithmetic Progression Question , whose 1st term is 23 and common difference is -3

Tn = a+ (n-1)d

-4 = 23 +(n-1)(-3)

n = 10 .
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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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11 Jan 2017, 08:09
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enigma123 wrote:
In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn= tn-1 - 3 for each n > 1. What is the value of n when tn = -4?

A. -1
B. 7
C. 10
D. 14
E. 20

In the given sequence, since we are given the first term, we can use that value to find the second term, and once we know the second term, we can use that value to find the third term, and so on.

t_1 = 23

t_2 = t_1 – 3 = 23 – 3 = 20

t_3 = t_2 – 3 = 20 – 3 = 17

t_4 = t_3 – 3 = 17 – 3 = 14

t_5 = t_4 – 3 = 14 – 3 = 11

t_6 = t_5 – 3 = 11 – 3 = 8

t_7 = t_6 – 3 = 8 – 3 = 5

t_8 = t_7 – 3 = 5 – 3 = 2

t_9 = t_8 – 3 = 2 – 3 = -1

t_10 = t_9 – 3 = -1 – 3 = -4

So n = 10.

Alternative Solution:

Notice that starting from the second term, each term is 3 less than the previous term, which makes the sequence an arithmetic sequence. In an arithmetic sequence, the nth term, a_n, can be found by using the formula a_n = a_1 + d(n – 1) in which a_1 is the first term and d is the common difference.

Since we are given t_n, we can modify the formula to t_n = t_1 + d(n – 1) in which t_1 = 23 and d = -3. So we have:

t_n = t_1 + d(n – 1)

-4 = 23 + (-3)(n – 1)

-27 = -3(n – 1)

9 = n – 1

10 = n

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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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08 Jun 2017, 16:44
Top Contributor
Attached is a visual that should help.
Attachments

Screen Shot 2017-06-08 at 4.33.09 PM.png [ 108.8 KiB | Viewed 28366 times ]

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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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05 Jul 2017, 19:18
Here is how I did it:

t1 = 23 and tn-1 = -3; therefore for every t you need to subtract 3

lets begin:

t1----> 23-3 = 20
t2----> 20-3 = 17
t3----> 17-3 = 14

when tn = 4...now subtract 4 from the 14

tn----> 14-4 = 10

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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn [#permalink]

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28 Feb 2018, 22:41
Hi All,

Questions that use "sequence notation" are relatively rare on Test Day (you'll probably see just 1), but the math behind the sequence is usually some fairly simple arithmetic (add, subtract, multiply, divide).

Here, we're given the first term in the sequence (23) and we're told that each term thereafter is 3 LESS than the preceding term. Once you understand how the sequence "works", in many cases, it's really easy to just "map out" the sequence. We're asked which term in the sequence equals -4.....

1st = 23
2nd = 20
3rd = 17
4th = 14
5th = 11
6th = 8
7th = 5
8th = 2
9th = -1
10th = -4

[Reveal] Spoiler:
C

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Re: In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn   [#permalink] 28 Feb 2018, 22:41
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