How do you find the nth term rule for #1/2,1,3/2,2,...#?

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Answer 1 · 2016-11-09T09:12:06

#nth. =1/2n#

#1/2,1,3/2,2,...#

firstly see what the terms increase by

difference #d=1-1/2=1/2#

so nth term is of the form

#nth.=1/2n+c#

use the first term and #n=1# to evaluate #c#

#1/2=1/2xx1+c#

#=>c=0#

#:.nth. =1/2n#

a quick mental check on the next few terms confirms the result.

Answer 2 · 2016-11-09T09:17:57

#T_n = 1/2n#

The sequence might be easier to recognise as an AP if the terms are written as improper fractions:

#T_n = 1/2, " "1," "1 1/2," " 2," " 2 1/2," " 3" "...#

An even better way to write this is in halves:

#T_n = 1/2, " "2/2," " 3/2, " "4/2," " 5/2...#

#" "n = 1, " " 2 ," " 3 , " "4 , " "5 .....#

It is then immediately obvious that each term is just #n/2#

Using the formula/ general rule:

The nth term is written as #T_n = a + d(n-1)#

All the values you need are available from the sequence.

#a = 1/2, " " d = 1/2" "# substitute into #T_n = a + d(n-1)#

#T_n= 1/2 + 1/2(n-1)" "larr# multiply out and simplify

#T_n = 1/2+1/2n-1/2#

#T_n = 1/2n#