What is the 8th term of the geometric sequence where a1 = 256 and a3 = 16?

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Answer 1 · 2018-03-09T12:42:10

#8# th term of the geometric sequence is #1/64#

Geometric sequence is # {a, ar ,ar^2...}; a and r # being first term

and common ratio respectively. #a_1=256 , a_ 3=16#

#n# th term is # a_n= a_1*r^(n-1) :. a_3=a_1*r^(3-1)=16# or

#256*r^2=16 or r^2= 16/256=1/16 :. r=1/4 #

#8# th term is # a_8= a_1*r^(n-1)=256(1/4)^(8-1)# or

# a_8=256(1/4)^7= 4^4/4^7=1/4^3=1/64#

#8# th term of the geometric sequence is #1/64# [Ans]