How do you write an nth term rule for #r=1/8# and #a_1=4#?

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Answer 1 · 2017-10-17T13:44:12

#a_n=4(1/8)^(n-1)#

presumably this is a geometric sequence question

in which case
for a GP

#a_1=#the first term#

#r=# the common ratio

we ahve

#a_1, a_1r, a_1r^2,...,a_1r^(n-1),..#

in this case we re given

#a_1=4#

#r=1/8#

so #a_n=4(1/8)^(n-1)#