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#### Series:

- If a
_{1}, a_{2}, a_{3}, ….., a_{n}. is a sequence, then the expression a_{1 }+ a_{2 }+ a_{3}+ …..+ a_{n}is called series.

#### Progression:

- It is not the case every time that the terms of sequence follow certain pattern so that a specific formula for the nth term of the sequence can be specified explicitly.
- If the terms of sequence follow certain pattern so that a specific formula for the nth term of the sequence can be specified explicitly, then the sequence is called the progression.

### Arithmetic Progression

- A sequence (t
_{n}) is said to be an arithmetic progression (A.P.) if t_{n + 1}– t_{n}= constant for all n ∈ N - The constant difference is called the common difference of the A.P. and is denoted by letter ‘d’. The first term is denoted by the letter ‘a’.

#### General Term or n^{th} Term of and Sum of an Arithmetic Progression:

- If ‘a’ is the first term and ‘d’ is the common difference of an A.P. then the n
^{th}term is given by

t_{n} = a + (n – 1)d

- If ‘a’ is the first term and ‘d’ is the common difference of an A.P. then the sum of the first n terms is given by

St_{n} = n/2(2a + (n – 1)d) = n/2(First term + Last term)

**Type – IA: To Find **t_{n }for Arithmetic Progression (A.P.).

_{n }for Arithmetic Progression (A.P.).

#### Algorithm:

- Write the first term = a, common difference = d
- the n
^{th}term of an A.P. is given by t_{n}= a + (n – 1)d - Substitute value of a and d in equation given in step 2
- Simplify R.H.S. to get answer.

#### Example – 01:

- Find n
^{th}term of an arithmetic progression (A.P.) 32, 28, 24, 20, ….. **Solution:**

Given A.P. is 32, 28, 24, 20, …..

First term = a = 32, common difference = d = 28 – 32 = – 4

the n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

∴ t_{n} = 32 + (n – 1)(-4) = 32 – 4n + 4 = 36 – 4n

**Ans:** The n^{th} term of the A.P. is (36 – 4n)

#### Example – 02:

- Find n
^{th}term of an arithmetic progression (A.P.) 4, 9, 14, 19, …….. **Solution:**

Given A.P. is 4, 9, 14, 19, ……..

First term = a = 4, common difference = d = 9 – 4 = 5

the n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

∴ t_{n} = 4 + (n – 1)(5) = 4 + 5n – 5 = 5n – 1

**Ans:** The n^{th} term of the A.P. is (5n – 1)

#### Example – 03:

- Find t
_{n}of an arithmetic progression (A.P.) 4, 14/3, 16/3, 6, …….. **Solution:**

Given A.P. is 4, 14/3, 16/3, 6, ……..

First term = a = 4, common difference = d = 14/3 – 4 = (14 – 12)/3 = 2/3

the n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

∴ t_{n} = 4 + (n – 1)(2/3) = 4 + 2n/3 – 2/3 = 2n/3 + 10/3 = (2n+ 10)/3

**Ans:** The n^{th} term of the A.P. is (2n+ 10)/3

**Type – IB: To Find Indicated Term of**** Arithmetic Progression (A.P.).**

#### Algorithm:

- Write the first term = a, common difference = d
- the n
^{th}term of an A.P. is given by t_{n}= a + (n – 1)d - Substitute value of a and d in equation given in step 2
- Simplify R.H.S. to get t
_{n}. - Substitute value of n the equation obtained in step 4 and simplify to get answer.

#### Example – 04:

- Find 24
^{th}term of an arithmetic progression (A.P.) 5, 8, 11, 14, …….... **Solution:**

Given A.P. is 5, 8, 11, 14, ……..

First term = a = 5, common difference = d = 8 – 5 = 3

the n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

∴ t_{n} = 5 + (n – 1)(3) = 5 + 3n – 3 = 3n + 2

∴ t_{24} = 3 x 24 + 2 = 72 + 2 = 74

**Ans:** The 24^{th} term of the A.P. is 74

#### Example – 05:

- Find 15
^{th}term of an arithmetic progression (A.P.) 21, 16, 11, 6, …….. **Solution:**

Given A.P. is 21, 16, 11, 6, ……..

First term = a = 21, common difference = d = 16 – 21 = -5

the n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

∴ t_{n} = 21 + (n – 1)(-5) = 21 – 5n + 5 = 26 – 5n

∴ t_{15} = 26 – 5 x 15 = 26 – 75 = – 49

**Ans:** The 15^{th} term of the A.P. is – 49

#### Example – 06:

- The 10
^{th}term of an arithmetic progression (A.P.) is 1 and 20^{th}term is – 29. Find the 3^{rd}term. **Solution:**

The n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

Given t_{10} = 1 and t_{20} = – 29

∴ t_{10} = a + (10 – 1)d = 1

∴ a + 9d = 1 ………… (1)

∴ t_{20} = a + (20 – 1)d = – 29

∴ a + 19d = – 29 ………… (2)

Subtracting equation (1) from (2) we get

10 d = -30

∴ d = – 3

Substituting in equation (1) we get

a + 9(-3) = 1

∴ a – 27 = 1

∴ a = 28

Now, t_{n} = a + (n – 1)d

∴ t_{3} = 28 + (3 – 1)(-3) = 28 – 6 = 22

**Ans:** The 3^{rd} term of the A.P. is 22

#### Example – 07:

- The 7
^{th}term of an arithmetic progression (A.P.) is 30 and 10^{th}term is 21. Find the 4^{th}term. **Solution:**

The n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

Given t_{7} = 30 and t_{10} = 21

∴ t_{7} = a + (7 – 1)d = 30

∴ a + 6d = 30 ………… (1)

∴ t_{10} = a + (10 – 1)d = 21

∴ a + 9d = 21 ………… (2)

Subtracting equation (1) from (2) we get

3 d = -9

∴ d = – 3

Substituting in equation (1) we get

a + 6(-3) = 30

∴ a – 18 = 30

∴ a = 48

Now, t_{n} = a + (n – 1)d

∴ t_{4} = 48 + (4 – 1)(-3) = 48 – 9 = 39

**Ans:** The 4^{th} term of the A.P. is 39

#### Example – 08:

- The 3
^{rd}term of an arithmetic progression (A.P.) is – 11 and 9^{th}term is – 35. Find the n^{th}term. **Solution:**

The n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

Given t_{3} = – 11 and t_{9} = – 35

∴ t_{3} = a + (3 – 1)d = – 11

∴ a + 2d = – 11 ………… (1)

∴ t_{9} = a + (9 – 1)d = – 35

∴ a + 8d = – 35 ………… (2)

Subtracting equation (1) from (2) we get

6 d = – 24

∴ d = – 4

Substituting in equation (1) we get

a + 2(-4) = – 11

∴ a – 8 = – 11

∴ a = = – 3

Now, t_{n} = a + (n – 1)d

∴ t_{n} = – 3 + (n – 1)(- 4) = – 3 – 4n + 4 = 1 – 4n

**Ans:** The n^{th} term of the A.P. is (1 – 4n)

#### Example – 09:

- Which term of the AP, : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer.
- Solution:

Given A.P. is 21, 18, 15, . . .

First term = a = 21, common difference = d = 18 – 21 = – 3

The n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

Given a + (n – 1)d = – 81

∴ 21 + (n – 1)(-3) = – 81

∴ (n – 1)(-3) = – 102

∴ n – 1 = 34

∴ n = 35

Thus – 81 is 35^{th} term

Given a + (n – 1)d = 0

∴ 21 + (n – 1)(-3) = 0

∴ (n – 1)(-3) = – 21

∴ n – 1 = 7

∴ n = 8

Thus 0 is 8^{th} term

#### Example – 10:

- Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .
**Solution:**

Given list. is 5, 11, 17, 23, . . .

First term = a = 5, common difference = d = 11 – 5 = 17 -11 = 23 – 17 = 6

The n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

Given a + (n – 1)d = 301

∴ 5 + (n – 1)(6) = 301

∴ (n – 1)(6) = 296

∴ n – 1 = 296/6 = 148/3

∴ n = 148/3 + 1 = 153/3

But n should be a positive integer but in this case it is fraction

Hence 301 is not a term of the list.

#### Example – 11:

- Determine the AP whose 3rd term is 5 and the 7th term is 9.
**Solution:**

The n^{th} term of an A.P. is given by t_{n} = a + (n – 1)d

Given t_{3} = 5 and t_{7} = 9

∴ t_{3} = a + (3 – 1)d = 5

∴ a + 2d = 5 ………… (1)

∴ t_{7} = a + (7 – 1)d = 9

∴ a + 6d = 9 ………… (2)

Subtracting equation (1) from (2) we get

4d = 4

∴ d = 1

Substituting in equation (1) we get

a + 2(1) = 5

∴ a = 3

Thus the A.P. is 3, 4, 5, 6, ….

**Ans:** The n^{th} term of the A.P. is (1 – 4n)

#### Example – 12:

- If for a sequence (t
_{n}), S_{n}= 4n^{2}– 3n, find t_{n}and show that the sequence is an A.P. **Solution:**

Given S_{n} = 4n^{2} – 3n ……….. (1)

S_{n – 1} = 4(n – 1)^{2} – 3(n – 1) = 4(n^{2} – 2n + 1) – 3n + 3 = 4n^{2} – 8n + 4 – 3n + 3

S_{n – 1} = 4n^{2} – 11n + 7 ……….. (2)

Subtracting equation (2) from (1)

t_{n} = S_{n} – S_{n – 1} = (4n^{2} – 3n) – (4n^{2} – 11n + 7)

t_{n} = 4n^{2} – 3n – 4n^{2} + 11n – 7 = 8n – 7

The n^{th} term of an A.P. is t_{n} = 8n – 7 ……….. (3)

∴ t_{n + 1} = 8(n + 1) – 7 = 8n + 8 – 7

∴ t_{n + 1} = 8n + 1 ……….. (4)

Subtracting equation (3) from (4) we get

t_{n + 1} – t_{n} = (8n + 1) – (8n – 7) = 8n + 1 -8n + 7 = 8

Thus the common difference is d = 8 which is the constant term

**Ans:** The sequence (t_{n}) is an A.P.

#### Example – 13:

- If for a sequence (t
_{n}), S_{n}= 2n^{2}+ 5n, find t_{n}and show that the sequence is an A.P. **Solution:**

Given S_{n} = 2n^{2} + 5n ……….. (1)

S_{n – 1} = 2(n – 1)^{2} + 5(n – 1) = 2(n^{2} – 2n + 1) + 5n – 5 = 2n^{2} – 4n + 2 + 5n – 5

S_{n – 1} = 2n^{2} + n – 3 ……….. (2)

Subtracting equation (2) from (1)

t_{n} = S_{n} – S_{n – 1} = (2n^{2} + 5n) – (2n^{2} + n – 3)

t_{n} = 2n^{2} + 5n – 2n^{2} – n + 3 = 4n + 3

The n^{th} term of an A.P. is t_{n} = 4n + 3 ……….. (3)

∴ t_{n + 1} = 4(n + 1) + 3 = 4n + 4 + 3

∴ t_{n + 1} = 4n + 7 ……….. (4)

Subtracting equation (3) from (4) we get

t_{n + 1} – t_{n} = (4n + 7) – (4n + 3) = 4n + 7 – 4n – 3 = 4

Thus the common difference is d = 4 which is the constant term

**Ans:** The sequence (t_{n}) is an A.P.

Science > Mathematics > Sequence and Series > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |

Error in first example of AP series please check them when anyone see that in starting that’s confused the viewer ..

Thank you

Thank you Sandeep. Corrections made