Activity 6

Prepared by Carla Benson, UMass Lowell

The three numbers, a, h and b form a harmonic sequence if their reciprocals, (1/a), (1/h), and (1/b) form an arithmetic sequence.

The number h is called the harmonic mean of a and b.

a .    Find the harmonic mean of  (1/6)and (1/12).

Use the fact that in an arithmetic sequence, each term differs from the one before it by the same amount.

Solve[(1/h) - 6 == 12 - (1/h), h]

b.  Express h in terms of a and b.

Solve[(1/h) - (1/a) == (1/b) - (1/h), h]

c.  Show that (1/a), (1/h)and(1/b)form an arithmetic sequence.

Check to see that the differences are the same:

Simplify[(((1/h) - (1/a)) /. 
          h -> ((2  a b)/(a + b)))]

Simplify[(((1/b) - (1/h)) /. 
          h -> ((2  a b)/(a + b)))]

As an alternative form of the check,use the formula for the (n^th) term of an arithmetic sequence.

(Subscript[c, n]) =  (Subscript[c, 1]) +  (n - 1) d

with d  =  (1/h) -  (1/a)

h = ((2  a b)/(a + b));

(Subscript[c, 1]) =  (1/a);
(Subscript[c, n_ ]) :=  
      Simplify[(1/a)  +  (n - 1) ((1/h) - (1/a))]

{(Subscript[c, 1]), (Subscript[c, 2]), (Subscript[c, 3])}

Comments

We use (Subscript[c, n]) in the formula for an arithmetic sequence instead of (Subscript[a, n]) to avoid confusion with the constant a.