# Harmonic Progressions Questions and Answers

## Harmonic Progression Questions and Answers in Aptitude

Harmonic Progression Questions and Answers are provided on this page for students to practice and get an idea how this topic is asked in the exam.

### General form of Harmonic Progression:

It is also a classification of real numbers in a way that any term in the order is the harmonic mean of its two fellow numbers.

Furthermore, if the converse of a sequence follows the rule of an arithmetic progression, then it is known as the harmonic progression.

It basically means that if a, a+d, a+2d, and so on is an A.P. then $\frac{1}{a}, \frac{1}{(a+d)}, \frac{1}{(a+2d)}$, and so on is known as H.P.

• The terms of a harmonic progression are always positive and non-zero. This is because the reciprocals of the terms must be defined, and division by zero is undefined.

• Unlike arithmetic and geometric progressions, harmonic progressions do not have a finite sum. The sum of the terms of a harmonic progression diverges, meaning it approaches infinity as you add more terms.

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1. Find the 6th and 9th term of the series 6, 4, 3, …

$\frac{12}{5}, \frac{4}{3}$

$\frac{12}{5}, \frac{4}{3}$

11.43%

$\frac{8}{5},\frac{7}{11}$

$\frac{8}{5},\frac{7}{11}$

8.57%

$\frac{4}{3},\frac{12}{7}$

$\frac{4}{3},\frac{12}{7}$

14.29%

$\frac{12}{7},\frac{6}{5}$

$\frac{12}{7},\frac{6}{5}$

65.71%

Consider $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}, ...... ∞$

Here $T_{2} – T_{1} = T_{3} – T_{2} = \frac{1}{12}$$\frac{1}{6}, \frac{1}{4}, \frac{1}{3}$ is an A.P.

6th term of this A.P. = $\frac{1}{6} + 5 \times \frac{1}{12} = \frac{1}{6} + \frac{5}{12} = \frac{7}{12}$,

And the 9th term = $\frac{1}{6} + 8 \times \frac{1}{12} = \frac{5}{6}$.

Hence the 9th term of the H.P. = $\frac{6}{5}$ and the 6th term = $\frac{12}{7}$

2. Find the harmonic mean of 2, 4, and 6

0.45

0.45

10.81%

1.34

1.34

13.51%

4.12

4.12

8.11%

3.27

3.27

67.57%

Harmonic Mean = $\frac{3abc}{ab +bc +ca}$

= $\frac{3\times 2\times 4\times 6}{2\times 4 + 4\times 6 + 6\times 2}$

= $\frac{144}{8 +24 +12} = 3.27$

3. Given the following frequency distribution of second year students of a particular college, calculate the harmonic mean.

 Age (Years) 15 16 17 18 19 Number Of Students 3 7 13 2 5

18

18

3.57%

19

19

10.71%

17

17

78.57%

16

16

7.14%

The given distribution is grouped data and the variable involved is the ages of first year students, while the number of students represents frequencies.

 Ages (Years) X Number of Students f f/x 15 3 0.2 16 7 0.4375 17 13 0.7647 18 2 0.1111 19 5 0.2632 Total ∑f = 30 ∑(f/x) = 1.7765

Harmonic mean
¯X=(∑f)/(∑(f/x)) = $\frac{30}{1.7765}$ = 16.8871 ≈17 years

4. Calculate the harmonic mean for the data given below:

 Marks 30-39 40-49 50-59 60-69 70-79 80-89 90-99 F 7 25 32 20 11 3 2

44

44

10.34%

54

54

72.41%

64

64

6.9%

74

74

10.34%

Calculations are as given below:

 Marks x f f/x 30 − 39 34.5 7 0.2029 40 − 49 44.5 25 0.5618 50 − 59 54.5 32 0.5872 60 − 69 64.5 20 0.3101 70 − 79 74.5 11 0.1477 80 −89 84.5 3 0.0355 90 − 99 94.5 2 0.0212 ∑f = 100 1.866

Harmonic Mean = $\frac{100}{1.866}$ =53.59 ≈54

5. If x, y, z are in h.p, then y is connected with x and z as :

$2\times\frac{1}{y} = \frac{1}{x} + \frac{1}{z}$

$2\times\frac{1}{y} = \frac{1}{x} + \frac{1}{z}$

59.52%

$2 \times\frac{1}{z} = \frac{1}{y} + \frac{1}{z}$

$2 \times\frac{1}{z} = \frac{1}{y} + \frac{1}{z}$

7.14%

$2 \times\frac{1}{x} = \frac{1}{x} + \frac{1}{y}$

$2 \times\frac{1}{x} = \frac{1}{x} + \frac{1}{y}$

16.67%

None of the mentioned

None of the mentioned

16.67%

If x, y, z are in h.p, then $\frac{1}{x} , \frac{1}{y} ,\frac{1}{z}$ will be in A.P. series and $\frac{1}{y}$ will be the AM of x, z.

6. For 2 number X.Y, Harmonic Mean among them will be?

$\frac{2y +2x}{3y}$

$\frac{2y +2x}{3y}$

6.98%

$\frac{2xy}{x+y}$

$\frac{2xy}{x+y}$

79.07%

$\frac{(x+y)}{2xy}$

$\frac{(x+y)}{2xy}$

2.33%

$\frac{2y}{(x+y)}$

$\frac{2y}{(x+y)}$

11.63%

Let z be the harmonic mean,

z = $\frac{2}{\frac{1}{x} + \frac{1}{y}} = \frac{2xy}{x+y}$

7. From the given data 5, 10,17,24,30 calculate H.M.

12.123

12.123

11.76%

11.526

11.526

73.53%

10.324

10.324

8.82%

11.932

11.932

5.88%

Explanation:

 X 1/X 5 0.2000 10 0.1000 17 0.0588 24 0.0417 30 0.0333

HM = $\frac{5}{0.4338} = 11.526$

8. If the sum of reciprocals of first 9 terms of an HP series is 81, find the 5th term of HP.

10

10

8.57%

9

9

17.14%

$\frac{1}{9}$

$\frac{1}{9}$

68.57%

$\frac{1}{8}$

$\frac{1}{8}$

5.71%

Now from the above HP formula, it is clear the reciprocals of first 9 terms will make an AP.

The sum of first 9 terms of an AP =$\frac{9}{2} [2a + (9 – 1) d]$  = 81

⇒ 2a + 8d = 18

⇒ a + 4d = 9

Now there are 2 variables, but a + 4d = $T_{5}$ in an AP series.

And reciprocal of 5th term of AP series will give the 5th term of corresponding HP series.

So, the 5th term of HP series is $\frac{1}{9}$

9. The 3rd term of an HP is 5 and the 6th term is 8. Find the maximum possible number of terms in H.P.

12

12

8.57%

9

9

5.71%

10

10

71.43%

11

11

14.29%

If a, a + d, a + 2d, a + 3d, ……. are in A.P.

Then $\frac{1}{a}, \frac{1}{a+d},\frac{1}{a+2d}, \frac{1}{a+3d}$, ....   are in H.P.

Now $\frac{1}{(a+2d)} = 5$  and $\frac{1}{(a+5d)} = 8$

Solving these two equations we get $a = \frac{1}{4}$

$d =\frac{-1}{40}$

Now a+10d = $(\frac{1}{4}) + (\frac{-10}{40}) = 0$

Hence the maximum terms that this H.P. can take is 10.

10. Find maximum partial sum of harmonic progression, if second and third terms are $\frac{1}{13}$ and $\frac{1}{10}$ respectively.

1.23

1.23

6.67%

1.63

1.63

76.67%

1.25

1.25

10%

1.09

1.09

6.67%

The terms of the HP are

$\frac{1}{16}, \frac{1}{13} , \frac{1}{10}, \frac{1}{7}, \frac{1}{4},\frac{1}{1},\frac{1}{-2}\frac{1}{-5}$

So the maximum partial sum is

$\frac{1}{16} + \frac{1}{13} + \frac{1}{10} + \frac{1}{7} +\frac{1}{4} + \frac{1}{1}$ = 1.63

11. The sixth term of H.P. is $\frac{9}{106}$ and seventh term is $\frac{9}{122}$. Find the sum of its second and fifth term.

$\frac{513}{3233}$

$\frac{513}{3233}$

9.68%

$\frac{11}{35}$

$\frac{11}{35}$

77.42%

$\frac{81}{63}$

$\frac{81}{63}$

9.68%

$\frac{897}{41}$

$\frac{897}{41}$

3.23%

If  a, a+d, a+2d… are in A.P. then

$\frac{1}{a}, \frac{1}{(a+d)}, \frac{1}{(a+2d)}$ … are in H.P.

Now $\frac{1}{a+5d} = \frac{9}{106} => a+5d = \frac{106}{9}$

And $\frac{1}{a+6d} = \frac{9}{122} => a+6d = \frac{122}{9}$

Solving these two equations, we get

a= $\frac{26}{9}$ and $d=\frac{16}{9}$

Now a+d = $\frac{14}{3}$ and a+4d = $\frac{10}{1}$

So the sum of second and fifth term of H.P is $\frac{3}{14} + \frac{1}{10} = \frac{11}{35}$

12. x , y ,z are said to be in harmonic progression if the reciprocals $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in a.p. find out the value of A for which 13,a,16 are in harmonic progression

12.23

12.23

9.09%

13.65

13.65

6.06%

14.34

14.34

75.76%

15.66

15.66

9.09%

As we know that 1/13, 1/a and 1/16 are in arithmetic progression.

or $\frac{1}{a}-\frac{1}{13} = \frac{1}{16}-\frac{1}{a}$

$\frac{-1}{13}-\frac{1}{16} = \frac{-1}{a}-\frac{1}{a}$

$\frac{-16-13}{208} = \frac{-2}{a}$

29a = 416

a = 14.34

13. A.M and G.M of two numbers is 13 and 12 respectively then H.M is _____

11/134

11/134

5.71%

13/144

13/144

8.57%

144/13

144/13

77.14%

21/167

21/167

8.57%

G²=A.H=12²=13XH

144/13=H

Harmonic mean=144/13

14. If (b+c), (c+a) and (a+b) are in H.P then $\frac{1}{a^{2}} , \frac{1}{b^{2}} , \frac{1}{c^{2}}$will be in

A.P

A.P

13.51%

H.P

H.P

70.27%

G.P

G.P

8.11%

None of the above

None of the above

8.11%

(b+c), (c+a), (a+b) are in h.p.

$\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ will be in A.P.

t₂-t₁=t₃-t₂

=$\frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}$

By solving the equation we get:

(b-a)(a+b)=(c-b)(b+c)

b²-a²=c²-b²

a²,b²,c² are in A.P.

$\frac{1}{a^{2}} , \frac{1}{b^{2}} , \frac{1}{c^{2}}$ are in H.P.

15. Find the nth term of H.P.$\frac{1}{2}, \frac{1}{7}, \frac{1}{12}, \frac{1}{17}$, … … ….

$\frac{1}{(n-3)}$

$\frac{1}{(n-3)}$

5.88%

(5n-3)

(5n-3)

8.82%

$\frac{1}{(5n-3)}$

$\frac{1}{(5n-3)}$

79.41%

(5n+3)

(5n+3)

5.88%

As we see the above series is in A.P.

So $a_{n}$=a+(n-1)d=2+(n-1)5=

5n-3

So nth term of h.p. = $\frac{1}{(5n-3)}$

16. Find the 10th term of the H.P.

1/3, 1/7, 1/11, 1/15,… … …

39/7

39/7

5.88%

1/39

1/39

79.41%

25/6

25/6

5.88%

25/7

25/7

8.82%

Here a₁₀=3+(10-1)4

=39

So 10th term will be 1/39.

17. Harmonic Mean of 1/a and 1/b is ______

$\frac{2}{a+b}$

$\frac{2}{a+b}$

75%

$\frac{a+b}{2}$

$\frac{a+b}{2}$

11.11%

$\frac{2}{ab}$

$\frac{2}{ab}$

5.56%

None of the above

None of the above

8.33%

H.P. Formula:

harmonic mean of a and b = $\frac{2ab}{a+b} = \frac{2}{a+b}$

18. The 7th term of $\frac{1}{2},\frac{1}{5},\frac{1}{8},\frac{1}{11},\frac{1}{14},\frac{1}{17}$ will be?

$\frac{1}{22}$

$\frac{1}{22}$

5.56%

$\frac{1}{21}$

$\frac{1}{21}$

13.89%

$\frac{1}{21}$

$\frac{1}{21}$

5.56%

None of the mentioned

None of the mentioned

75%

As we know the above series is in AP thus

a = 2 + (7-1)3 = 20

$\frac{1}{20}$ is h term of the series.

19. Find the harmonic mean of the following data {5, 9, 7, 10, 8, 5}?

6.66

6.66

3.23%

7

7

9.68%

6.82

6.82

80.65%

8

8

6.45%

We have: {5, 9, 7, 10, 8, 5}

H.M = $\frac{6}{\frac{1}{5} + \frac{1}{9} + \frac{1}{7} + \frac{1}{10} + \frac{1}{8} + \frac{1}{5}} = 6.82$

20. Find harmonic mean, if A.M is 100 and G.M is 65?

29.40

29.40

2.94%

42.25

42.25

76.47%

32.40

32.40

8.82%

44.78

44.78

11.76%

$GM^{2} = AM \times HM$

HM = $\frac{65^{2}}{100} = 42.25$

21. Find the harmonic mean of 2 numbers whose G.M. and A.M. is 90 and 75?

103

103

9.38%

116

116

9.38%

108

108

75%

117

117

6.25%

$GM^{2} = AM \times HM$

HM = $\frac{90^{2}}{75} = 108$

22. The harmonic mean of 17,22,13,26 will be ?

14.32

14.32

3.23%

16.76

16.76

9.68%

17.90

17.90

16.13%

21.05

21.05

70.97%

Formula is

$H.M = \frac{n}{\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}}+...\frac{1}{a_{n}}}$

for number 17,22,13,26 harmonic mean is -
$\frac{4}{( \frac{1}{17} + \frac{1}{22} + \frac{1}{13} + \frac{1}{26} )}$
=$\frac{4}{0.05+0.04+0.07+0.03}$
harmonic mean = 21.05

23. If there are two sets with 150 and 132 as harmonic mean and comprising 25 and 12 observations then the combined harmonic mean....?

7.85

7.85

10.34%

9.53

9.53

79.31%

5.82

5.82

3.45%

None of the above

None of the above

6.9%

The mean of the reciprocals of the terms = 1/150 and 1/132.

Thus, the total of the reciprocals of the terms in set i=1/150×25=1/6
and in set ii=1/132×12=1/11

The sum of the reciprocals of all 37 components is 1/6+1/11=17/66.

Hence their mean is 17/66÷37=9.53.

24. Find the harmonic mean of two numbers 19 and 57?

25.31

25.31

10%

27.34

27.34

10%

28.50

28.50

70%

24.21

24.21

10%

HM =2xy/(x+y)

So x=19 and y=57

H.M=(2*19*57)/(19+57)

H M=2166/76=28.50

25. Find the reciprocal of  harmonic mean of two numbers 18 and 37?

1/15.31

1/15.31

6.45%

1/17.43

1/17.43

9.68%

1/24.21

1/24.21

77.42%

1/20.21

1/20.21

6.45%

HM =2xy/(x+y)

So x=18 and y=37

H.M=(2*18*37)/(18+37)

H.M=1332/55=24.21.

Reciprocal of this=1/24.21

26. A.M and G.M of two numbers is 27 and 26 respectively then H.M is _____

22.54

22.54

5.88%

23.54

23.54

23.53%

25.03

25.03

64.71%

21.77

21.77

5.88%

G²=A.H=26²=27*H

676/27=H

Harmonic mean=25.03

27. Find the 11th term of the H.P.

1/4, 1/8, 1/12, 1/16,… … …

39/7

39/7

5.71%

1/44

1/44

85.71%

25/6

25/6

5.71%

25/7

25/7

2.86%

Here  $a_{11}$=4+(11-1)4

=44

So 11th term will be 1/44.

28. If p, q, r are in h.p, then q is connected with p and r as :

2(1⁄q) = (1⁄p + 1⁄r)

2(1⁄q) = (1⁄p + 1⁄r)

57.14%

2(1⁄r) = (1⁄q + 1⁄r)

2(1⁄r) = (1⁄q + 1⁄r)

14.29%

2(1⁄p) = (1⁄p + 1⁄q)

2(1⁄p) = (1⁄p + 1⁄q)

11.43%

None of the mentioned

None of the mentioned

17.14%

1p1⁄q, 1r will be in a.m. series and 1⁄q will be the A.M of p, r.

29. For 2 numbers U, V, Harmonic Mean among them will be?

(2v+2u )/3v

(2v+2u )/3v

5.88%

2uv/(u+v)

2uv/(u+v)

82.35%

(u+v)/2uv

(u+v)/2uv

2.94%

2v/(u+v)

2v/(u+v)

8.82%

Let w be the harmonic mean, 2w = 1u + 1v,

W= $\frac{2uv}{(u+v)}$

30. Find maximum partial sum of harmonic progression, if second and third terms are $\frac{1}{13}$ and $\frac{1}{10}$ respectively.

1.23

1.23

6.9%

1.63

1.63

79.31%

1.25

1.25

3.45%

1.09

1.09

10.34%

The terms of the HP are

$\frac{1}{16}, \frac{1}{13} , \frac{1}{10}, \frac{1}{7}, \frac{1}{4},\frac{1}{1},\frac{1}{-2}\frac{1}{-5}$

So the maximum partial sum is

$\frac{1}{16} + \frac{1}{13} + \frac{1}{10} + \frac{1}{7} +\frac{1}{4} + \frac{1}{1}$ = 1.63

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