Question 1165029
<pre>
4/5,2/3,4/7,... n=12

First we get the nth term of the reciprocals, 

5/4,3/2,7/4,...

which is an arithmetic sequence with first term 5/4 and common difference
3/2-5/4 = 6/4-5/4 = 1/4
7/4-3/2 = 7/4-6/4 = 1/4

a<sub>n</sub> = a<sub>1</sub>+(n-1)d

{{{a[n]=5/4+(n-1)expr(1/4)}}}

{{{a[n]=5/4+n*expr(1/4)-1*expr(1/4)}}}

{{{a[n]=5/4+n*expr(1/4)-1/4}}}

{{{a[n]=4/4+expr(1/4)n}}}

{{{a[n]=(4+n)/4}}}

So when n=12

{{{a[n]=(4+12)/4}}}

{{{a[n]=(16)/4}}}

{{{a[n]=4}}}

So the 12th term of the given harmonic sequence is the reciprocal of 4.

Answer:  1/4

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Checking with a TI-84

Type in 4/5 
press MATH ENTER ENTER

Type in 

(,  2ND, (-), x<sup>-1</sup>, +, 1, /, 4, ), x<sup>-1</sup>, MATH, ENTER   
 
note: the (-) refers to the 'negative' key at the bottom just below the 3.  

Then press ENTER over and over to generate the terms up to the 12th one:

  1   2   3   4   5   6    7   8    9  10   11  12 
4/5,2/3,4/7,1/2,4/9,2/5,4/11,1/3,4/13,2/7,4/15,1/4 

Edwin</pre>