How do you find the Nth term in the geometric sequence where the first term is 3 and the fourth term is #6sqrt2#?

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Answer 1 · 2015-12-31T14:44:53

#a_N = 3*2^((N-1)/2)#

The general term of a geometric sequence can be written:

#a^n = a r^(n-1)#

where #a# is the initial term and #r# the common ratio.

In our case we are given #a_1 = 3# and #a_4 = 6sqrt(2)#

So we find:

#r^3 = (ar^3)/(ar^0) = a_4/a_1 = (6sqrt(2))/3 = 2sqrt(2) = (sqrt(2))^3#

So (assuming the geometric sequence is Real):

#r = root(3)(sqrt(2)^3) = sqrt(2)#

and

#a_n = 3 * (sqrt(2))^(n-1) = 3 * 2^((n-1)/2)#