How do you write the first five terms of the geometric sequence #a_1=81, a_(k+1)=1/3a_k# and determine the common ratio and write the nth term of the sequence as a function of n?

1 Answer
Shwetank Mauria
Mar 27, 2017

First five terms are #{81,27,9,3,1]#, common ratio is #1/3# and #n^(th)# term #a_n=3^(5-n)#

Explanation:

In a geometric series common ratio #r# is the ratio of term to its immediately preceding term. Here we have a term #a_k# and as next term is #a_(k+1)# and #a_(k+1)/a_k=1/3#,

we have #r=1/3#

AS given first term as #a_1# and common ratio as #r#, the #n^(th)# term #a_n=a_1xxr^(n-1)#. Now given first term #a_1=81# and #r=1/3#

#n^(th)# term #a_n# is #81xx1/3^(n-1)#

= #3^4/3^(n-1)=3^(4-n+1)=3^(5-n)#

first five terms are #{81,81xx1/3=27,27xx1/3=9,9xx1/3=3,3xx1/3=1}#

i.e. #{81,27,9,3,1}#

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