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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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