How To Solve Harmonic Progression Questions Quickly

How to Solve Harmonic Progression questions quickly

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 1/a1, 1/a2, 1/a3……1/an
where 1/a1 = the first term, and  d = the common difference.

Type 1: nth term of an HP: tn = 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.1/10

C.1/6

D.1/5

Solution

Reciprocals of first 11 terms will be AP

Therefore, Sn =  n/2[2a + (n − 1) d]

S11=110, n =11

110 =   11/2[2a + (11 − 1) d]

110 = 11/2 * [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 = 1/10

Correct option: B

Question 2

Find the infinite sum of series 1/(32-4) + 1/(42-4) + 1/(52-4) …….

Options:

A. 0.45

B.0.69

C.0.87

D.2

Solution

In the given series,

S= 1/(32-4) + 1/(42-4) + (1/52-4) + …..

S= 1/(32 – 22) + 1/(42-22) + 1/(52– 22) + …..

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

S= 1/[(3-2) (3 +2)] + 1/[(4-2) (4+2)] + 1/[(5-2) (5+2)] + …..

S = ¼(1-1/5) + ¼(1/2 – 1/6) + ¼(1/3 – 1/7)

S = ¼ * 1.833

S = 0.45

Correct option: A

Question 3.

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

Options:

  A. 12/5 and 13/9

B. 12/9 and 13/5

C. 12/5 and 12/9

D. 12/5 and 12/7

Solution

Consider the series in the form of 1/6, 1/4, 1/3

We know that T2 – T1 = T3 – T2 = 1/12

As 1/6, 1/ 4, 1/3 is an AP, Thus 4th term of the AP = 1/ 6 + 3 * 1/ 12 = 5/12

Similarly the 8th term of the series = 1/ 6 + 7 * 1/ 12 = 9/12

Thus the 4th term of an AP = 12/5 and 8th term = 12/9 respectively.

Correct option: C

how to solve harmonic progression quickly

Type 2: Harmonic mean of the series: HM = n/{1/a1+1/a2…1/an}

Question 1.

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

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 = 4 / (1/2 + 1/3 + 1/6)

Average speed = 4 / 1

Average speed = 4 km/hr

Correct option: C

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 =n/{1/a1+1/a2…1/an}

HM =3{1/5+1/10+1/15}

HM = 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

Options:

 A. A ≥ G ≥ H

B. A < G > H

C. A < G ≥ H

D. A < G < H

Solution

Arithmetic mean (A) = 4 +6 / 2 = 5

Geometric mean (G) = √4*6

Harmonic mean (H) = 2/( 1/ 4 + 1/ 6)= 4.8

Hence it is shown that A ≥ G ≥ H

Correct option: A

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