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]