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About HP:

As a part of Sequence and series, it’s much interesting. Earlier you have read about Arithmetic Progression (A.P), Harmonic progression series has the same order only the change is that it follows the reciprocal of Arithmetic Progression.

We can define Harmonic Progression as the reciprocal of the sequence as followed by the sequence of A.P. H.P is also known as Harmonic Sequence.

Formally defined as “A sequence a_n of real numbers is a Harmonic Progression (HP) if any term in the sequence is the harmonic mean of its two neighbours”.

If a, a+d, a+2d, a+3d, …. are in Arithmetic Progression (A.P) then \(\frac{1}{a}\), \(\frac{1}{a+d}\), \(\frac{1}{a+2d}\), \(\frac{1}{a+3d}\), … are in Harmonic Progression (H.P).

Harmonic Mean:

Let a, b and c are series in H.P. Then as we know, \(\frac{1}{a}\), \(\frac{1}{b}\), \(\frac{1}{c}\) form an A.P.

Then \(\frac{2}{b}\)= \(\frac{1}{a}\) + \(\frac{1}{c}\). Hence, b = \(\frac{2ac}{a+c}\). This ‘b’ is called as the harmonic mean of a and c.

Important points to learn about HP:

As n^th term of AP = a + (n-1)d,
So n^th term of HP = \(\frac{1}{(a + (n-1)d)}\)
Let there are two positive real numbers a & b.
Then, Arithmetic Mean x Harmonic Mean = (Geometric Mean)²Simply, AM x HM = GM
The significant relation between all three types of Mean are
AM > GM > HM

Mathematical Assumption in HP:

If we need to assume three numbers in a H.P. then they should be

\(\frac{1}{a-d}\), \(\frac{1}{a}\), \(\frac{1}{a+d}\) .

Four numbers assumption in H.P. are

\(\frac{1}{a-3d}\), \(\frac{1}{a-d}\), \(\frac{1}{a+d}\), \(\frac{1}{a+3d}\)

Five convenient numbers in H.P. are

\(\frac{1}{a-2d}\), \(\frac{1}{a-d}\), \(\frac{1}{a}\), \(\frac{1}{a+d}\), \(\frac{1}{a+2d}\),

It will be easy for you in most of the questions solve it by converting into A.P.

Sum of Terms in a Harmonic Progression:

Not exactly but approximately given by

S_n≈ \(\frac{1}{d}\)ln(\(\frac{2a+(2n-1)d}{2a-d}\))

Examples related to HP:

Q. If the 4^th term of HP is 8 and 10^th term is 16. Then find the maximum number of terms.

Solution:

As we know that reciprocal of HP is AP.
a+3d = \(\frac{1}{8}\)
a+9d = \(\frac{1}{16}\)
from these two equation a= \(\frac{5}{32}\), d = \(\frac{-1}{96}\)
Now, a + (n-1)d = 0 so that n0 more terms are possible.

a + 15d = 0
The given series can take only 15 terms before vanishing to 0.

So max. no. of terms = 15

Q.If sum of 3 consecutive terms in an HP is \(\frac{1}{6}\) and sum of their reciprocals is \(\frac{1}{4}\). What are the three numbers?

Solution:

(a-d) + a + (a+d) =\(\frac{1}{4}\)
3a = \(\frac{1}{4}\)
a = \(\frac{1}{12}\)
Now,
\(\frac{1}{12-d}\)+\(\frac{1}{12}\)+\(\frac{1}{12+d}\) = \(\frac{-1}{6}\)

Thus, d =\(\frac{31}{2}\)
Rest is clear.

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