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Collected some where online, i have no quick solution for it

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Manager
Joined: 02 Mar 2008
Posts: 203
Concentration: Finance, Strategy
Collected some where online, i have no quick solution for it [#permalink]

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28 Jul 2008, 06:36
Collected some where online, i have no quick solution for it

b1=10; b2=20
bn=[b(n-2) + 2*b(n-1)]/3
...
b3=(b1+2*b2)/3

Find the 15th item of the sequence

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Senior Manager
Joined: 19 Mar 2008
Posts: 345

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28 Jul 2008, 08:01
AlbertNTN wrote:
Collected some where online, i have no quick solution for it

b1=10; b2=20
bn=[b(n-2) + 2*b(n-1)]/3
...
b3=(b1+2*b2)/3

Find the 15th item of the sequence

I cannot think of any shortcut. Any expert can advise?
Manager
Joined: 02 Mar 2008
Posts: 203
Concentration: Finance, Strategy

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29 Jul 2008, 09:04
hi there, anyone can help?
Director
Joined: 27 May 2008
Posts: 531

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29 Jul 2008, 10:02
AlbertNTN wrote:
hi there, anyone can help?

can you provide options. backsolving or plugging can be a good approach in this question.
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1345

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29 Jul 2008, 19:20
1
First, notice that the definition of the nth term is just a weighted average of the two previous terms. That is, the nth term will always be between the (n-1)st term and the (n-2)nd term, and will be twice as far from the (n-2)nd term as it is from the (n-1)st term. In other words, if a, b, c are three consecutive terms in the sequence, we will always have:

|c-a| = 2|c-b| (in other words, the distances will be in a 2 to 1 ratio)
and either a < c < b or b < c < a

So if b_1 = 10, b_2 = 20, then:

b_3 = 10 + (2/3)(20 - 10) [b_3 is 10 plus 2/3rds of the distance between 10 and 20]
b_4 = 10 + (2/3)(10) + 1/9(10) [the distance from b_2 to b_3 is 1/3 of the distance from b_1 to b_2]
b_5 = 10 + (2/3)(10) + (1/9)(10) - (1/27)(10)
b_6 = 10 + (2/3)(10) + 1/9(10) - (1/27)(10) + (1/81)(10)

and so on.

In general, for i > 2,
b_i = 50/3 + SUM_(k from 2 to i-2) [(-1)^k * 10 * (1/3^k)]

This sum converges very quickly. If we work out the sum over an infinite number of terms, we'll get an excellent approximation of b_15. Rewriting:

b_i = 50/3 + 10*SUM_(k from 2 to i-2) [(-1)/3]^k

Let S = SUM_(k from 2 to i-2) [(-1)/3]^k
S = (1/3^2 - 1/3^3) + (1/3^4 - 1/3^5) + (1/3^6 - 1/3^7) + ...
S = 2/3^3 + 2/3^5 + 2/3^7 + ...
S = 2(1/3^3 + 1/3^5 + 1/3^7 + ...)
S = 2[1/3^3 + (1/3^2)(1/3^3 + 1/3^5 + 1/3^7 + ...)]
S = 2(1/3^3 + (1/3^2)(S/2))
S = 2/27 + S/9
8S/9 = 2/27
S = 1/12

(or you could just use a formula for the sum of an alternating series here).

So limit(i --> infinity) b_i = 50/3 + 10*(1/12) = 100/6 + 5/6 = 17.5

b_15 will be *very* close to 17.5. Working out the exact value of b_15 is a bit annoying, but you can do it following the method above; I find that it's equal to:

(82*3^6 + 10)(20/3^13)

Still, I'd use the answer choices in a question like this - and would never expect to see such a question on the GMAT!
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Director
Joined: 27 May 2008
Posts: 531

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29 Jul 2008, 22:11
good explanation Ian.
A quick help from MS excel gives the answer 17.5 (well almost )

n b_n
1 10
2 20
3 16.66666667
4 17.77777778
5 17.40740741
6 17.53086420
7 17.48971193
8 17.50342936
9 17.49885688
10 17.50038104
11 17.49987299
12 17.50004234
13 17.49998589
14 17.50000470
15 17.49999843
Manager
Joined: 02 Mar 2008
Posts: 203
Concentration: Finance, Strategy

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30 Jul 2008, 07:34
systematic explanation. Thanks. I got fr my friend's hand-written doc, and no A, B, C, D, E choices also...
VP
Joined: 17 Jun 2008
Posts: 1290

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03 Aug 2008, 05:57
AlbertNTN wrote:
Collected some where online, i have no quick solution for it

b1=10; b2=20
bn=[b(n-2) + 2*b(n-1)]/3
...
b3=(b1+2*b2)/3

Find the 15th item of the sequence

at n=3 ,b3=[b+4*b]/3=5b/3
5b/3= (10+40)/3 => b=10

now b15= [10*(13)+ (20*(14))]/3=410/3
This is the quickest i can work out

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Re: Sequence   [#permalink] 03 Aug 2008, 05:57
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Collected some where online, i have no quick solution for it

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