How do you write the expression for the nth term of the geometric sequence #a_3=16/3, a_5=64/27, n=7#?

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Answer 1 · 2018-05-14T00:39:50

#a_n=2^(n+1)/3^(n-2)#
So:
#a_7=256/243#

#a_3=16/3,a_5=64/27#

#a_5=( 16 * 2 * 2 )/( 3 * 3 * 3 )=a_3 *2/3*2/3#

So: #a_(n+1)=a_n * 2/3#

That will mean:
#a_3=a_2*2/3 rArr a_2=a_3/ (2/3)=(16/2) / (3/3) rArr a_2=8/1#

So, we have:
#a_2=2^3/3^0, a_3=2^4/3^1, a_5=2^6/3^3 #

We can deduct:
#a_n=2^(n+1)/3^(n-2)#

So:
#a_7=2^8/3^5=256/243#