Harmonic Progressions: Solved Examples

Q.1. If the sum of reciprocals of first 11 terms of an HP series is 110, find the 6^th term of HP.

Explanation: Now from the above HP formulae, it is clear the reciprocals of first 11 terms will make an AP. The sum of first 11 terms of an AP = [2a + (11 – 1) d] 11/2 = 110
⇒ 2a + 10d = 20 ⇒ a + 5d = 10
Now there are 2 variables, but a + 5d = T₆ in an AP series. And reciprocal of 6^th term of AP series will give the 6^th term of corresponding HP series. So, the 6^th term of HP series is 1/10

Q.2. Find the 4th and 8th term of the series 6, 4, 3, …

Explanation: Consider 1/6, /14, 1/3, ...
Here T₂ – T₁ = T₃ – T₂ = 1/12
Therefore 1/6, 1/4, 1/3 is in A.P.
4th term of this Arithmetic Progression = 1/6 + 3 × 1/12 = 1/6 + 1/4 = 5/12,
Eighth term = 1/6 + 7 × 1/12 = 9/12.
Hence the 8th term of the H.P. = 12/9 = 4/3 and the 4th term = 12/5.

Q.3. The 2nd term of an HP is 40/9 and the 5th term is 20/3. Find the maximum possible number of terms in H.P.

Explanation : If a, a + d, a + 2d, a + 3d, ……. are in A.P. then are in H.P.
Now and
Solving these two equations we get and
Now
Hence the maximum terms that this H.P. can take is 10.

Q.4. If the first two terms of a harmonic progression are 1/16 and 1/13, find the maximum partial sum?

Answer: Option A

Explanation : The terms of the HP are
So the maximum partial sum is

Q.5. The second term of an H.P. is and the fifth term is . Find the sum of its 6^th and the 7^th term.

Explanation: If a, a + d, a + 2d, a + 3d, ……. are in A.P. then are in H.P.
Now
and
Solving these two equations, we get and
Now
And
So the sum of the 6^th and the 7^th term of H.P. is

Q.6. If the sixth term of an H.P. is 10 and the 11th term is 18 Find the 16th term.

Answer: Option C