How do you write an nth term rule for #a_3=10# and #a_6=300#?

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Answer 1 · 2018-02-23T03:10:16

#a_n=1.0359*3.107^(n-1)#

I'm assuming this is a geometric sequence...

The general formula for an arithmetic sequence is

#a_n=a_1*r^(n-1)# with #a_1# as the first term and #r# as the common ratio.

Plug in:

#a_3=10=a_1*r^(3-1)#
#a_6=300=a_1*r^(6-1)#

#10=a_1*r^(2)#
#300=a_1*r^(5)#

Divide:

#300/10=(a_1*r^(5))/(a_1*r^(2))#

#30=(cancel(a_1)*r^(5))/(cancel(a_1)*r^(2))#

#30=r^3#

#root3(30)=r or 30^(1/3)#

#r~~3.107#

Then find #a_1#:

#10=a_1*9.653#

#a_1~~1.0359#

Plug in the information into the general formula:

#a_n=1.0359*3.107^(n-1)#

Note: This is not exact as I rounded somewhere.