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PS- GMAT Prep

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brcinsf
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PS- GMAT Prep [#permalink] New post   21 Jun 2006, 22:34
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Ok, I am stumped. How does one solve this question? Any clear & concise explanation will be appreciated.

2 Days till GMAT,
Tru



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Re: PS- GMAT Prep [#permalink] New post   22 Jun 2006, 01:00
t1 = 23
t2 = t1-3 = 23-3 = 20
t3 = t2 - 3 = t1 - 3 - 3 = t1 - 2 (3) = 23 - 2 (3) = 17
t4 = t3 - 3 = t2 - 3 - 3 - 3 = t1 - 3 (3) = 23 - 3 (3) = 14

we can establish a relationship as under:

tn = t1 - (n-1) (3)
-4 = t1- (n - 1) (3)
(n -1) (3) = 23 +4
(n-1) (3) = 27
n = 9 +1 = 10
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Re: PS- GMAT Prep [#permalink] New post   22 Jun 2006, 04:09
Answer is 10
Every term of sequence is previous term minus three, sequence is
23-20-17-14-11-8-5-2-(-1)-(-4)
(-4) is the 10th term
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Re: PS- GMAT Prep [#permalink] New post   22 Jun 2006, 04:12
We have t(n)=t(1)-(n-1)*3
So if t(n)=-4 => 23-3*(n-1)=4 => n=10
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Re: PS- GMAT Prep [#permalink] New post   22 Jun 2006, 09:33
You can quickly go from t1 = 23 and list out the values
{23, 20, 17, 14, 11, 8,5,2,-1, -4} i.e. 10
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Re: PS- GMAT Prep [#permalink] New post   22 Jun 2006, 10:03
It's all about the pattern! Thank you everyone.

-Tru
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Re: PS- GMAT Prep [#permalink] New post   22 Jun 2006, 16:02
tn = t1 + (n-1)d; question specifies d = -3

=> -4 = 23 - 3n + 3 => n = 30.
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Re: PS- GMAT Prep [#permalink] New post   22 Jun 2006, 18:15
brcinsf wrote:
It's all about the pattern! Thank you everyone.

-Tru


Use nth term of a sequence = first_term+(n -1)(difference_in_terms)
(or usually remembered as Tn = a+(n-1)d

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

-4 = 26 -3n => n =10



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Re: PS- GMAT Prep [#permalink]
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