How To Solve Arthmetic And Harmonic Progression Quickly

How to Solve AP and GP and HP Questions

In this Page you will learn How to Solve AP and GP and HP Questions Quickly through different types of Questions.This Topic is very important from Examination Point of View. That’s why This will helps in different Examinations.

Definition

AP

An Arithmetic progression is a sequence of numbers in which each term is derived from the preceding term by adding or subtracting a fixed number called the common difference “d”.

GP

A Geometric progression is a sequence in which each term is derived by multiplying or dividing the preceding term by a fixed number called the common ratio. 

HP

A series of terms is known as a HP series when their reciprocals are in arithmetic progression. 

 

How to Solve AP and GP and HP

Type 1: AP questions

Question 1.

Find the first term of the AP series in which 10th term is 6 and 18th term is 70.
Options:

  1. 76
  2. – 76
  3. 66
  4. – 66

Solution:

10th term = (a + 9d) = 6….(1)

18th term = (a + 17d) = 70 ……. (2)

On solving equation 1 and 2

We get, d = 8

Put the value of d in equation 1

(a + 9d) = 6

a + 9 x 8 = 6

a + 72 = 6

a = -66

Correct option: 4

Question 2.

Find the nth term of the series 3, 8, 13, 18,…,

Options:

  1. 2(2n+ 1)
  2. 5n + 2
  3. 5n – 2
  4. 2(2n – 1)

Solution:

The given series is in the form of AP.
first term a = 3

common difference d = 5
We know that, nth term = tn = a + (n-1)d
Therefore, tn = 3 + (n-1) 5

= 3 + 5n – 5

= 5n – 2

Correct option: 3

Question 3.

The series 28, 25,……. -29 has 20 terms. Find out the sum of all 20 terms?

Options:

  1.  -10
  2. -12
  3. 10
  4. 12

Solution:

a =28, d= -3 (25 – 28), l = -29, n = 20

Sum of all n-terms = Sn = \frac{n}{2}(a+l)

S20 = \frac{20}{2} (28 + (-29))
S20 = -10

Correct option: 1

Type 2: GP questions

Question 1.

Find the sum of the following infinite G. P. \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}…….

Options:

  1. \frac{1}{3}
  2. \frac{2}{3}
  3. \frac{1}{5}
  4. \frac{1}{2}

Solution:

a = \frac{1}{3}, r = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3}

Required sum = \frac{a}{(1-r)}

= \frac{\frac{1}{3}}{(1-\frac{1}{3})}

= \frac{\frac{1}{3}}{\frac{2}{3}}

= \frac{1}{2}

Correct option: 4

Question 2.

Find the G. M. between \frac{4}{25} and \frac{196}{125}

Options:

  1. \frac{28}{5}
  2. \frac{28}{25}
  3. \frac{8}{25}
  4. \frac{14}{5}

Solution:

Geometric mean \sqrt{ab}

GM = \sqrt{\frac{4}{25}\times \frac{196}{25}}

GM = \frac{28}{25}

Correct option: 2

Question 3.

Find the number of terms in the series 1, 3, 9 , ….19683

Options:

  1. 10
  2. 8
  3. 6
  4. 7

Solution:

In the given series,

a1 = 1, r = \frac{3}{1} = 3, an =19683

=> 19683 = 1 x (3n-1)

=>19683 = 3n-1

=>39 = 3n-1

=>9 = n-1

n = 10

Correct option: 1

Type 3: HP questions

Question 1:

If the 6th term of H.P. is 10 and the 11th term is 18. Find the 16th term.

Options:

  1. 90
  2. 110
  3. 85
  4. 100

Solution:

6th term = a + 5d = \frac{1}{10}……(1)

11th term = a + 10d = \frac{1}{18}……(2)

On solving equation 1 and 2 we get, d = \frac{-2}{225}

Put value of d in equation 1

a + 5d = \frac{1}{10}

a + 5 \times \frac{-2}{225} = \frac{1}{10}

a = \frac{13}{90}

Now, 16th term = a + 15d = \frac{13}{90} + 15 \times \frac{- 2}{225}

= \frac{13}{90} – \frac{30}{225}

= \frac{1}{90}

Therefore 16th term = 90

Correct option: 1

Question 2.

Find the Harmonic mean of 6, 12, 18

Options:

  1. 10.12
  2. 9.62
  3. 9.81
  4. 8.10

Solution:

We know that,

HM = \frac{n}{s}

where , s = (\frac{1}{a})+(\frac{1}{b})+(\frac{1}{c})

s=(\frac{1}{6})+(\frac{1}{12})+(\frac{1}{18})

=(\frac{11}{36})

HM = \frac{n}{s}=\frac{3}{(\frac{11}{36})}

HM = 3\times \frac{36}{11}

HM = \frac{108}{11}

HM=9.81

Correct option: 3

Question 3.

What is the relation between AM, GM, and HM?

Options:

  1. AM x HM = GM2
  2. AM / HM = GM
  3. AM + HM = GM2
  4. AM – HM = GM2

Solution:

AM = \frac{a+b}{2}

GM = \sqrt{ab}

HM = \frac{2ab}{a+b}

Therefore AM x HM = GM2

\frac{a+b}{2} \times  \frac{2ab}{a+b}= ab

Correct option: 1

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7 comments on “How To Solve Arthmetic And Harmonic Progression Quickly”


  • SHOURYA

    Let P = {2, 3, 4, ………. 100} and Q = {101, 102, 103, ….. 200}. How many elements of Q are there such that they do not have any element of P as a factor ?

    what is the method to find the prime no. between the numbers as soon as possible ?


  • Sadabrata

    type 3 harmonic progression. Please explain
    Question 2.
    Find the Harmonic mean of 6, 12, 18

    Options:

    10.12
    10.9
    10.06
    6.10
    Solution:
    We know that,

    HM = 3/0.298

    HM = 10.06

    Correct option: C


    • Gunasri

      formula= n/(1/a1 +1/a2 + 1/a3)
      6,12,18 we have 3 numbers
      so n=3
      then 3/(1/6 + 1/12 +1/18)


      • Naveen kumar

        ormula= n/(1/a1 +1/a2 + 1/a3)
        6,12,18 we have 3 numbers
        so n=3
        then 3/(1/6 + 1/12 +1/18)
        after using this formula we get 9.868 answer

        Find the Harmonic mean of 6, 12, 18
        Options:
        10.12
        10.9
        10.06
        6.10
        Solution:
        We know that,
        HM = 3/0.298
        HM = 10.06 ——-???


    • kirti

      Please explain
      type 3 harmonic progression.
      Question 2.
      Find the Harmonic mean of 6, 12, 18 Options: 10.12
      10.9
      10.06
      6.10
      Solution:
      We know that, HM = 3/0.298 HM = 10.06 Correct option: C