 # Harmonic Progressions: Solved Examples

Q.1. If the sum of reciprocals of first 11 terms of an HP series is 110, find the 6th term of HP.
1. 1/5
2. 1/10
3. 2/7
4. 1/20
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 6th term of AP series will give the 6th term of corresponding HP series. So, the 6th term of HP series is 1/10
Q.2. Find the 4th and 8th term of the series 6, 4, 3, …
1. 12/5 and 4/3
2. 7/10 and 5/7
3. 1/7 and 3/4
4. 20/11 and 9/8
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.
1. 11
2. 10
3. 9
4. 12
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?
1. 1.63
2. 1.13
3. 1.89
4. 2.2
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 6th and the 7th term.
1. 177/1547
2. 145/1457
3. 513/3233
4. 117/3123
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 6th and the 7th 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.
1. 30
2. 75
3. 90
4. 80
Explanation: Here And Solving these two equations, we get and We have 