The 3rd term of a harmonic progression is 15 and the 9th term is 6. Find the 11th term.

In a Harmonic Progression (HP), the reciprocals of the terms form an Arithmetic Progression (AP).

Given that the 3rd term of the HP is 15, the 3rd term of the corresponding AP is 1/15.

Similarly, the 9th term of the HP is 6, so the 9th term of the corresponding AP is 1/6.

We can find the common difference (d) of the AP using the formula:

d = (a_n - a_m) / (n - m)

where a_n and a_m are the nth and mth terms of the AP, and n and m are their respective positions.

Substituting the given values:

d = (1/6 - 1/15) / (9 - 3) = (5/90) / 6 = 5/540 = 1/108

The nth term of an AP is given by a_n = a + (n - 1)d

We know that the 3rd term of the AP is 1/15, so we can write:

1/15 = a + (3 - 1)(1/108)

Solving for a (the first term of the AP), we get:

a = 1/15 - 2/108 = 6/108 - 2/108 = 4/108 = 1/27

Now we can find the 11th term of the AP using the formula a_n = a + (n - 1)d:

a_11 = 1/27 + (11 - 1)(1/108) = 1/27 + 10/108 = 4/108 + 10/108 = 14/108 = 7/54

Remember that the terms of the HP are the reciprocals of the terms of the AP, so the 11th term of the HP is 1/(7/54) = 54/7 = 7.71 (approx).