# How To Solve Harmonic Progression Questions Quickly

## How to Solve Harmonic Progression Problems Quickly

A Harmonic Progression or Harmonic Sequence is a sequence or progression of real numbers formed by taking the reciprocals of an Arithmetic Progression or Arithmetic Sequence Here are some of the best way for How to Solve Harmonic Progression Problems effectively.

After Reading This Page you will Learn How to Solve Harmonic Progression Problems .In this Page also different types of Harmonic Progression Problems is given.

### Definition for HP

• A harmonic progression is a sequence of real numbers formed by taking the reciprocals of an Arithmetic progression.
• An HP is represented in the form $\frac{1}{a_{1}}, \frac{1}{a_{2}} ,\frac{1}{a_{3}}……\frac{1}{a_{n}}$

where

$\frac{1}{a_{1}}$= the first term,

d = the common difference taken from AP = d

### Type 1: nth term of an HP : $a_{n} = \frac{1}{a+(n-1)d}$

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

Options:

A. 10

B. $\frac{1}{10}$

C. $\frac{1}{6}$

D. $\frac{1}{5}$

Solution    Reciprocals of first 11 terms of an HP will be AP
Therefore, $s_{n} =\frac{n}{2}[2a+(n-1)\times d]$

$S_{n}$ = 110
n = 11
$110 = \frac{11}{2}[2a + (11 − 1) d]$
$110 = \frac{11}{2} \times [2a + 10d]$
220 = 22a + 110d
22a + 110d = 220
a + 5d = 10

which is the 6th term of the AP series

Therefore, the 6th term in HP = $\frac{1}{10}$

Correct option: B

Question 2 Find the infinite sum of series $\frac{1}{(3^{2}-4)} + \frac{1}{(4^{2}-4)} + \frac{1}{(5^{2}-4)}$ …….

Options:

A. 0.45

B. 0.69

C. 0.87

D. 2

Solution     In the given series,

$\frac{1}{(3^{2}-4)} + \frac{1}{(4^{2}-4)} + \frac{1}{(5^{2}-4)}$

$\frac{1}{(3^{2}-2^{2})} + \frac{1}{(4^{2}-2^{2})} + \frac{1}{(5^{2}-2^{2})}$

We know, that a2-b2 = (a-b) (a+b)

S = $\frac{1}{(3-2)(3+2)} + \frac{1}{(4-2)(4+2)} +\frac{1}{(5-2)(5+2)}$ …..

S = $\frac{1}{4}(1-\frac{1}{5}) + \frac{1}{4}(\frac{1}{2} – \frac{1}{6}) + \frac{1}{4}(\frac{1}{3} – \frac{1}{7})$

S = $\frac{1}{4}\times 1.833$

S = 0.45

Correct option: A

Question 3. Identify the 4th and 8th term of the series 6, 4, 3….

Options:

A. $\frac{12}{5}$ and $\frac{13}{9}$

B. $\frac{12}{9}$ and $\frac{13}{5}$

C. $\frac{12}{5}$ and $\frac{12}{9}$

D. $\frac{12}{5}$ and $\frac{12}{7}$

Solution     Consider the series in the form of $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}$

We know that $T_{2} – T_{1} = T_{3} – T_{2} = \frac{1}{12}$

As $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}$ is an AP, Thus 4th term of the AP = a +3d = $\frac{1}{ 6} + 3 \times \frac{1}{12} = \frac{5}{12}$

Similarly the 8th term of the series = a +7d = $\frac{1} {6} + 7 \times \frac{1}{12} = \frac{9}{12}$

Thus the 4th term of an AP = $\frac{12}{5}$ and 8th term = $\frac{12}{9}$ respectively.

Correct option: C

### Type 2: Harmonic mean of the series: HM = $\mathbf{\frac{n} {\frac{1}{a_{1}} + \frac{1}{a_{2}} ……\frac{1}{a_{n}}}}$

Question 1. Aarti walked first one-third of the distance at a speed of 2 km/hr. The next one-third of the distance was covered by running at the speed of 3km/hr. The last one-third of the distance was covered by cycling at the speed of 6 km/hr. Find the average speed for the whole journey covered by Aarti?

Options:

A. 5 km/hr

B. 6 km/hr

C. 4 km/hr

D. 3 km/hr

Solution    According to the question, the distance covered is same in all the three cases.

Therefore the average speed = HM of 2, 3, and 6

Average speed = $\frac{3}{ (\frac{1}{2} + \frac{1}{3} + \frac{1}{6})}$

Average speed = $\frac{3}{1}$

Average speed = 3 km/hr

Correct option: D

Question 2. Find the Harmonic mean of 5, 10 ,15

Options:

A. 4.5

B. 6.5

C. 8.33

D. 6.4

Solution      We know that,

HM = $\frac{n}{\frac{1}{a_{1}} + \frac{1}{a_{2}}…\frac{1}{a_{n}}}$

HM = $\frac{3}{\frac{1}{5}+\frac{1}{10}+\frac{1}{15}}$

HM = $\frac{3}{0.36}$

HM = 8.33

Correct option: C

Question 3. Find the correct option for A, G , H for the elements 4 and 6 . where A is Arithmetic Mean ,G is Geometric Mean & H is Harmonic Mean.

Options:

A. A ≥ G ≥ H

B. A < G > H

C. A < G ≥ H

D. A < G < H

Solution      Arithmetic mean (A) = $\frac{4 +6}{2} = 5$

Geometric mean (G) = $(4\times 6)^{\frac{1}{2}} = 4.89$

Harmonic mean (H) = $\frac{2}{( \frac{1}{4} + \frac{1}{6})}$ = 4.8

Hence it is shown that A ≥ G ≥ H

Correct option: A