How do you find nth term rule for #a_2=5# and #a_4=1/5#?

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Answer 1 · 2016-08-16T20:13:12

There are two possibilities:

#a_n = 25(1/5)^(n-1)#

#a_n = -25(-1/5)^(n-1)#

Assuming that this is a geomtric sequence, since that's the topic under which it is posted...

The general formula for the #n#th term of a geometric sequence is:

#a_n = ar^(n-1)#

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

In our example, we find:

#r^2 = (ar^3)/(ar^1) = a_4/a_2 = (1/5)/5 = 1/25#

Hence #r = +-sqrt(1/25) = +-1/5#

If #r=1/5# then #a = (ar)/r = a_2/r = 5/(1/5) = 25#

If #r = -1/5# then #a = (ar)/r = a_2/r = 5/(-1/5) = -25#