If a, b and c are in harmonic progression with b as their harmonic mean then,
This is obtained as follows. Since a, b and c are in harmonic progression, 1/a, 1/b and 1/c are in arithmetic progression. Then,
This can be written as
On cross multiplication we obtain
2ac=b(a + c)
|That is, b
The second proposition we are going to look at in this part is: If A, G and H are the arithmetic, geometric and harmonic means respectively between two given quantities a and b then G2 = AH. The explanation is given below.
We know that the arithmetic mean of a and b is and it is given that this equals to A.
|Similarly G2 = ab and H =
|The product of AH =
||= ab. This we observe is equal to G2.
That is, G2 = AH, which says that G is the geometric mean between A and H.
Insert two harmonic means between 4 and 12.
We convert these numbers into A.P. They will be 1/4 and 1/12. Including the two arithmetic means we have four terms in all. We are given the first and the fourth terms. Thus,
T0 = a = 1/4 and
T4 = a + 3d = 1/12
Substituting the value of a = 1/4 in T4, we have
1/4 + 3d = 1/12
3d = 1/12 - 1/4 = - 1/6
d = -1/18
Using the values of a and d, we obtain T2 and T3.
T2 = a + d = 1/4 + (-1/18)
= 1/4 - 1/18 = 7/36
T3 = a + 2d = 1/4 + 2.(-1/18)
= 1/4 - 2/18
= 1/4 - 1/9
The reciprocals of these two terms are 36/7 and 36/5.
Therefore, the harmonic series after the insertion of two means will be 4, 36/7, 36/5 and 12.