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
Difficult Math Problem #120 - Sequences
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- jayhawk2001
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a4 = b3 - a3800guy 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
= (b2 + a2) - (b2 - a2)
= 2a2
= 2 (b1 - a1)
= 2(k - 1)
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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)
=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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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)--------------------
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)--------------------