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    Given: 1, ½, 1/3, ¼…   Observe the denominators of each term in the sequence. Try to get the reciprocals. Do these form a sequence? If so, what type of sequence?   A sequence whose reciprocals form an arithmetic sequence is called a harmonic sequence.   Example1:   1, ½, 1/3, ¼…   The reciprocal form an arithmetic sequence   In the harmonic sequence 2/3, ½, 2/5, 1/3, 2/7…   We can say that ½ is the harmonic means between 2/3 and 2/5; ½, 2/5 and 1/3 are the harmonic between 2/3 and 2/7.   2/3+1/2+2/5+… is a harmonic series.   Do you think the harmonic sequence has similar properties as those of an arithmetic sequence   Example 2:   Find the 12th term of the harmonic sequence 1/9, 1/12, 1/15...   Note that the reciprocal forms an arithmetic sequence so we may first find the 12th term of the harmonic sequence which is 1/42   Example 3       Insert two harmonic means between 6 and 3/2. We first find the arithmetic means between 1/6 and 2/3. Since we have to insert two terms, then we have n= 4, a 1 = 1/6 and a 4 = 2/3   Using the formula a n = a 1 +(n-1)d, we can find the common difference.   a 4 = a 1 +(n-1)d   2/3= 1/6+(4-!)d   d=1/6   So the arithmetic means are:   a 2 = a 1 +d a 3 =a 2 +d   =1/6+1/6 =1/3+1/6   =1/3 =1/2   The reciprocals of these two terms result in the harmonic means between 6 and 3/2 which are 3 and 2   Example 4:   Find the sum of the harmonic series 3/2+6/7+3/5+...6/19. What did you find out when you explored the sum, we need to find the number   of terms in this sequence. We compute   n by using the formula a n =a 1 +(n-1)d.   We have a 1 =2/3, a n = 19/6 and d=7/6-2/3 or 1/2     an=a 1 +(n-1)d   so, 19/6=2/3=(n-1)1/2   1/2n=19/6-2/3+1/2   There are two missing terms, a 4  and a 5.   a 4 =a 3 +d a 5 =a 4 +d   =5/3+1/2 =13/6+1/2   =13/6 =8/3   We now have the complete terms of the harmonic series, 3/2+6/7+3/5+6/13+3/8+6/19.   S n = 3/2+6/7+3/5+6/13+3/8+6/19   =3(1/2+2/7+1/5+2/13+1/8+2/19)   = 94737 / 69160  
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