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

Question

b1=10; b2=20 bn= /3 ... b3=(b1+2*b2)/3 Find the 15th item of the sequence

judokan · Answer

I cannot think of any shortcut. Any expert can advise?

AlbertNTN · Answer

hi there, anyone can help?

durgesh79 · Answer

can you provide options. backsolving or plugging can be a good approach in this question.

IanStewart · Answer

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!

durgesh79 · Answer

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

AlbertNTN · Answer

systematic explanation. Thanks. I got fr my friend's hand-written doc, and no A, B, C, D, E choices also...

spriya · Answer

at n=3 ,b3= /3=5b/3 5b/3= (10+40)/3 => b=10 now b15= /3=410/3 This is the quickest i can work out

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards