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tough sequences

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by joanjgonzalez » Tue Dec 16, 2008 8:05 am
for an infinite sequence of numbers a1, a2, a3,....an for all n>1 a(n) = a(n-1)+4 if n is odd and a(n)=a(n-1)-1 if n is even. What is the value of a(1) if a(34)=68?

a. 10
b. 11
c. 12
d. 13
e. 21

imo is [spoiler]e

thanks!!!!
[/spoiler]
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by dmateer25 » Tue Dec 16, 2008 8:37 am
a(34) = a(34-1) - 1
68 = a(33) - 1
69 = a(33)

a(33) = a(33-1) + 4
69 = a(32) + 4
65 = a(32)


It is a pattern

Every a(odd) will increase by 1. Every a(even) will decrease by 4.

so we are going from 33 to 1

So there are 33 numbers. 17 are odd and 16 are even.

17 x 1 = 17
16 x -4 = -64


Add these 2 numbers to 68 to come up with a(1)

a(1) = 68 + 17 - 64
a(1) = 21

I will go with E
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by mrsmarthi » Tue Dec 16, 2008 6:09 pm
Ans is E. Here is an alternate solution.

Try to get couple of terms in the reverse order.
Given a(34) = a(33) - 1 = 68. ==> a(33) = 69
a(33) = a(32) + 4 ==> 69 = a(32) + 4 ==> a(32) = 65
a(32) = a(31) -1 ==> 65 = a(31) - 1 ==> a(31) = 66

Consider all the odd terms a(1), a(3), a(5) ......a(31), a(33).The series will be as follows

a(1), a(3)......66,69.

We have in total 17 odd terms. This is an Arithmetic Sequence. Applying the formula of nth term we should be getting a(1)

n = 17
d = 3
17th Term = 69.

Applying the formula for the nth term in the AP series, a + (n-1)d = nth term, we have

69 = a(1) + (17-1) 3
69 = a(1) + 48
a(1) = 21.
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