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## Answers

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, a1= 1/6 and a4= 2/3

Using the formula an= a1+(n-1)d, we can find the common difference.

a4= a1+(n-1)d

2/3= 1/6+(4-!)d

d=1/6

So the arithmetic means are:

a2= a1+d a3=a2+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 an=a1+(n-1)d.

We have a1=2/3, an= 19/6 and d=7/6-2/3 or 1/2

an=a1+(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, a4 and a5.

a4=a3+d a5=a4+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.

Sn= 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