# How To Solve Arthmetic And Harmonic Progression Quickly

## Definition and Solve Quickly

### 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 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 * 8 = 6

a + 72 = 6

a = -66

### 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

### 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 = n (a+l)/2

S20 = 20 (28 + (-29)) / 2
S20 = -10

## Type 2: GP questions

### Question 1.

Find the sum of the following infinite G. P. 1/3, 1/9, 1/27, 1/81…….

Options:

1. 1/3
2. 2/3
3. 1/5
4. 1/2

#### Solution:

a = 3, r = 1/9/1/3 = 1/3

Required sum = a/(1-r)

= 1/3 / (1-1/3)

= 1/3 / 2/3

= ½

### Question 2.

Find the G. M. between 4/25 and 196/25

Options:

1. 28/5
2. 28/25
3. 8/25
4. 14/5

#### Solution:

Geometric mean √ab

GM = √4/25 * √196/25

GM = 2/5 * 14/5

GM = 28/25

### 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 = 3/1 = 3, an =19683

=

19683 = 1* (3n-1)

19683 = 3n-1

39 = 3n-1

9 = n-1

n = 10

## 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 = 1/10……(1)

11th term = a + 10d = 1/18……(2)

On solving equation 1 and 2 we get, d = -2/225

Put value of d in equation 1

a + 5d = 1/10

a + 5 * -2/225 = 1/10

a = 13/90

Now, 16th term = a + 15d = 13/ 90 + 15 * – 2/55

= 13/90 – 30/225

= 1/90

Therefore 16th term = 90

### Question 2.

Find the Harmonic mean of 6, 12, 18

Options:

1. 10.12
2. 10.9
3. 10.06
4. 6.10

We know that,

HM = 3/0.298

HM = 10.06

### Question 3.

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

Options:

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

#### Solution:

AM = a+b/2

GM = √ab

HM = 2ab/a+b

Therefore AM * HM = GM2

a+b/2 * 2ab/a+b = ab