How do you write the nth term rule for the sequence #1/2,3,11/2,8,21/2,...#?

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Answer 1 · 2016-08-01T23:34:56

#T_n = (5n)/2 -2#

First decide whether it is an AP or GP.
Subtract consecutive terms.

#11/2-3 =5 1/2-3 = 2 1/2#
#3-1/2 = 2 1/2#

There is the same difference, so it is an AP.

We have the first term, and d.
Sub into the general term.

#T_n = a+ (n-1)d = 1/2 + 5/2(n-1)#

#T_n = 1/2 + (5n)/2 - 5/2#

#T_n = (5n)/2 -2#