Step-by-step explanation:
⇒12)It is an arithmetic sequence.
d=2-1=3-2=4-3=1
a(n) = a +(n-1)d
a(n) = 1+(n-1)1
The next three terms:
a(6) = 1+(6-1)1=6
a(7) = 1+(7-1)1=7
a(8) = 1+(8-1)1=8
⇒13)It is an arithmetic sequence.
d=0-3=-3-0=-6+3=-3
a(n) = a +(n-1)d
a(n) = 3+(n-1)-3
The next three terms:
a(5) = 3+(5-1)-3=-9
a(6) = 3+(6-1)-3=-12
a(7) = 3+(7-1)-3=-15
⇒14)It is <u>not </u>an arithmetic sequence.
⇒15) a(50) = 10 +(50-1)5
=<u>255</u>
<u>I hope this helps</u>
<u />
Not necessarily.
and
may be linearly dependent, so that their span forms a subspace of
that does not contain every vector in
.
For example, we could have
and
. Any vector
of the form
, where
, is impossible to obtain as a linear combination of these
and
, since
unless
and
.
If you mean how do you get x by it's you divide both sides by 8, to cancel out the multiplication between x and 8, meaning that it would end up being x=7, sorry if that didn't answer your question correctly.
