3 years ago

Let a=1,-3,2 and b= -1,3,-2 find a-3b

1 answer:

7 0

A=<1,-3,2> and b=<-1,3,-2>

a-3b=<1,-3,2>-3<-1,3,-2>
a-3b=<1,-3,2>-<3(-1),3(3),3(-2)>
a-3b=<1,-3,2>-<-3,9,-6>
a-3b=<1-(-3),-3-9,2-(-6)>
a-3b=<1+3,-12,2+6>
a-3b=<4,-12,8>

Answer: a-3b=<4,-12,8>

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The 12th term of a sequence is 41 and the 15th term is 140. What is the nth term?

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An= mth term.
an=a₁+(n-1)*d

a₁₂=41
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a₁₂=41
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With the equiations (1) and (2) build a system of equations
a₁+11d=41
a₁+14d=140

we solve it.
-(a₁+11d=41)
a₁+14d=140
--------------------
3d=99 ⇒d=99/3=33

a₁+11d=41
a₁+(11*33)=41
a₁+363=41
a₁=41-363=-322

an=a₁+(n-1)*d
an=-322+(n-1)*33
an=-322+33n-33
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an=-355+33n

To check:
a₁₂=-355+33*12=-355+396=41
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