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Difficult Math Problem #120 - Sequences

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Re: Difficult Math Problem #120 - Sequences

by jayhawk2001 » Wed Apr 18, 2007 8:02 pm
800guy wrote:Sequence A and B. a1=1, b1=k. an=b(n-1)-a(n-1) bn=b(n-1)+a(n-1). What is a4=?

from difficult math problems doc, oa coming after people answer with questions
a4 = b3 - a3
= (b2 + a2) - (b2 - a2)
= 2a2
= 2 (b1 - a1)
= 2(k - 1)
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by Cybermusings » Thu Apr 19, 2007 2:13 am
a4 = a(4-1)+b(4-1)
=a3+b3
Now a3=a(3-1)-b(3-1)
and b3=a(3-1)+b(3-1)
a3+b3 =a2-b2+a2+b2
=2a2
a2=a(2-1)-b(2-1)
=a1-b1
=1-k
so 2a2=2(1-k)
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OA

by 800guy » Fri Apr 20, 2007 8:43 am
OA:

a2 = k-1 ; b2 = k+1

a3= (k+1)-(k-1) = 2 ; b3 = (k+1)+(k-1) = 2k

a4 = 2k - 2 = 2(k-1) = 2(b1-a1)
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by rajeshvellanki » Sat Apr 21, 2007 8:27 am
long mehtod:

a1=1;b1=k

a2=b1-a1
==> a2=k-1

b2=b1+a1
==>b2=k+1

a3=b2-a2
a3=(k+1)-(k-1)
==>a3=2

b3=b2+a2
b3=(k+1)+(k-1)
==>b=2k

k now

a4=b3-a3
a4=(2k)-(2)
==>a4=2(k-1)


so ----------------a4=2(k-1)--------------------
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