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# 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.

**Type 1: ****AP questions**

### Question 1.

**Find the first term of the AP series in which 10 ^{th} term is 6 and 18^{th} term is 70. **

Options:

- 76
- – 76
- 66
- – 66

#### Solution:

10^{th} term = (a + 9d) = 6….(1)

18^{th} 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 n ^{th} term of the series 3, 8, 13, 18,…,**

Options:

- 2(2n+ 1)
- 5n + 2
- 5n – 2
- 2(2n – 1)

#### Solution:

The given series is in the form of AP.

first term a = 3

common difference d = 5

We know that, n^{th} term = t_{n} = a + (n-1)d

Therefore, t_{n} = 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:

- -10
- -12
- 10
- 12

#### Solution:

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

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

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

S_{20} = -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:

- \frac{1}{3}
- \frac{2}{3}
- \frac{1}{5}
- \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:

- \frac{28}{5}
- \frac{28}{25}
- \frac{8}{25}
- \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:

- 10
- 8
- 6
- 7

#### Solution:

In the given series,

a_{1} = 1, r = \frac{3}{1} = 3, a_{n }=19683

=> 19683 = 1 x (3^{n-1})

=>19683 = 3^{n-1}

=>3^{9} = 3^{n-1}

=>9 = n-1

n = 10

Correct option: 1

**Type 3: HP questions**

### Question 1:

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

Options:

- 90
- 110
- 85
- 100

#### Solution:

6^{th} term = a + 5d = \frac{1}{10}……(1)

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

= \frac{13}{90} – \frac{30}{225}

= \frac{1}{90}

Therefore 16^{th} term = 90

Correct option: 1

### Question 2.

**Find the Harmonic mean of 6, 12, 18**

Options:

- 10.12
- 9.62
- 9.81
- 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:

- AM x HM = GM
^{2} - AM / HM = GM
- AM + HM = GM
^{2} - AM – HM = GM
^{2}

#### Solution:

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

GM = \sqrt{ab}

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

Therefore AM x HM = GM^{2}

Correct option: 1

**Read also: **

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 ?

Kindly drop your query on 9711936107, then our mentor will resolve all your queries

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

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)

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 ——-???

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

These are very useful and easy. Tq prepinsta