# How To Solve Harmonic Progression Questions Quickly

## How to Solve Harmonic Progression Problems Quickly

### How to Solve HP Questions Quickly

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

Here are some of the best way for how to solve harmonic progression questions effectively.

### 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

### Related Banners

Get PrepInsta Prime & get Access to all 200+ courses offered by PrepInsta in One Subscription

## Get over 200+ course One Subscription

Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others