What is the 25th term of the sequence, where a_1=3 3,7,11,15,19
1 answer:
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The nth term is
an=a1(r)^(n-1)
an=1(2)^(n-1)
a1=1
r=2
the sum of a geometric seequence is
a1=1
r=2
we want to find
(since we minus 1, the highest exponet is 9 so add 1 to make it correct)
an=a1(r)^(n-1)
an=1(2)^(n-1)
a1=1
r=2
the sum of a geometric seequence is
a1=1
r=2
we want to find
