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

- A sequence is a function whose domain is the set N of natural numbers.
- Sequence is generally denoted by a and its image is denoted by a(n) or t(n). The terms of sequence are represented by a
_{1}, a_{2}, a_{3}, ….., a_{n }or t_{1}, t_{2}, t_{3}, ….., t_{n} - A sequence whose range is subset of R (set of real numbers) is called a real sequence.
- Population of human beings or bacteria at different times form a sequence. The amount of money deposited in a bank, over a number of years form a sequence. Depreciated values of certain commodity occur in a sequence.
- Sequences have important applications in many human activities.
- A sequence containing finite number of terms is called a finite sequence.
- A sequence is called infinite, if it is not a finite sequence.

### Type – IA: To Find First Three/ Five Terms of Sequence

#### Example – 01:

- Give first three terms of the sequence defined by a
_{n}= n/(n^{2}+ 1) **Solution:**

Given a_{n} = n/(n^{2} + 1)

By substituting n = 1, 2, 3 we get first three terms

a_{1} = 1/(1^{2} + 1) = 1/2

a_{2} = 2/(2^{2} + 1) = 2/5

a_{3} = 3/(3^{2} + 1) = 3/10

**Ans:** The first three terms are 1/2, 2/5 and 3/10.

#### Example – 02:

- Give first four terms of the sequence defined by a
_{n}= n^{2}– n + 1 **Solution:**

Given a_{n} = n^{2} – n + 1

By substituting n = 1, 2, 3, 4 we get first four terms

a_{1} = 1^{2} – 1 + 1 = 1 – 1 + 1 = 1

a_{2} = 2^{2} – 2 + 1 = 4 – 2 + 1 = 3

a_{3} = 3^{2} – 3 + 1 = 9 – 3 + 1 = 7

a_{4} = 4^{2} – 4 + 1 = 16 – 4 + 1 = 13

**Ans:** The first four terms are 1, 3, 7, 11

#### Example – 03:

- Give first three terms of the sequence defined by a
_{n}= 2n + 5 **Solution:**

Given a_{n} = 2n + 5

By substituting n = 1, 2, 3 we get first three terms

a_{1} = 2(1) + 5 = 2 + 5 = 7

a_{2} = 2(2) + 5 = 4 + 5 = 9

a_{3} = 2(3) + 5 = 6 + 5 = 11

**Ans:** The first three terms are 7, 9, 11.

#### Example – 04:

- Give first three terms of the sequence defined by a
_{n}= (n – 3)/4 **Solution:**

Given sequence is a_{n} = (n – 3)/4

By substituting n = 1, 2, 3 we get first three terms

a_{1} = (1 – 3)/4 = -2/4 = -1/2

a_{2} = (2 – 3)/4 = -1/4

a_{3} = (3 – 3)/4 = 0/4 = 0

**Ans:** The first three terms are -1/2, -1/4 and 0.

#### Example – 05:

- Give first five terms of the sequence defined by a
_{n}= n(n + 2) **Solution:**

Given a_{n} = n(n + 2)

By substituting n = 1, 2, 3, 4, 5 we get first five terms

a_{1} = 1(1+2) = 1(3) = 3

a_{2} = 2(2+2) = 2(8) = 8

a_{3} = 3(3+2) = 3(5) = 15

a_{4} = 4(4+2) = 4(6) = 24

a_{5} = 5(5+2) = 5(7) = 35

**Ans:** The first five terms are 3, 8, 15, 24, 35.

#### Example – 06:

- Give first five terms of the sequence defined by a
_{n}= n/(n + 1) **Solution:**

Given a_{n} = n/(n + 1)

By substituting n = 1, 2, 3, 4, 5 we get first five terms

a_{1} = 1/(1+1) = 1/2

a_{2} = 2/(2+1) = 2/3

a_{3} = 3/(3+1) = 3/4

a_{4} = 4/(4+1) = 4/5

a_{5} = 5/(5+ 1) = 5/6

**Ans:** The first five terms are 1/2, 2/3, 3/4, 4/5, and 5/6.

#### Example – 07:

- Give first five terms of the sequence defined by a
_{n}= 2^{n} **Solution:**

Given a_{n} = 2^{n}

By substituting n = 1, 2, 3, 4, 5 we get first five terms

a_{1} = 2^{1} = 2

a_{2} = 2^{2} = 4

a_{3} = 2^{3} = 8

a_{4} = 2^{4} = 16

a_{5} = 2^{5} = 32

**Ans:** The first five terms are 2, 4, 8, 16, 32.

#### Example – 08:

- Give first five terms of the sequence defined by a
_{n}= (2n – 3)/6 **Solution:**

Given a_{n} = (2n – 3)/6

By substituting n = 1, 2, 3, 4, 5 we get first five terms

a_{1} = (2 x 1 – 3)/6 = (2 – 3)/6 = -1/6

a_{2} = (2 x 2 – 3)/6 = (4 – 3)/6 = 1/6

a_{3} = (2 x 3 – 3)/6 = (6 – 3)/6 = 3/6 = 1/2

a_{4} = (2 x 4 – 3)/6 = (8 – 3)/6 = 5/6

a_{5} = (2 x 5 – 3)/6 = (10 – 3)/6 = 7/6

**Ans:** The first five terms are -1/6, 1/6, 1/2, 5/6, and 7/6.

#### Example – 09:

- Give first five terms of the sequence defined by a
_{n}= (-1)^{n – 1 }5^{n + 1} **Solution:**

Given a_{n} = (-1)^{n – 1 }5^{n + 1}

By substituting n = 1, 2, 3, 4, 5 we get first five terms

a_{1} = (-1)^{1 – 1 }5^{1 + 1} = (-1)^{0 }5^{2} = 1 x 25 = 25

a_{2} = (-1)^{2 – 1 }5^{2 + 1} = (-1)^{1 }5^{3} = – 1 x 125 = – 125

a_{3} = (-1)^{3 – 1 }5^{3 + 1} = (-1)^{2 }5^{4} = 1 x 625 = 625

a_{4} = (-1)^{4 – 1 }5^{4 + 1} = (-1)^{3 }5^{5} = -1 x 3125 = – 3125

a_{5} = (-1)^{5 – 1 }5^{5 + 1} = (-1)^{4 }5^{6} = 1 x 15625 = 15625

**Ans:** The first five terms are 25, – 125, 625, – 3125, 15625.

#### Example – 10:

- Give first five terms of the sequence defined by a
_{n}= n(n^{2}+ 5)/4 **Solution:**

Given a_{n} = n(n^{2} + 5)/4

By substituting n = 1, 2, 3, 4, 5 we get first five terms

a_{1} = 1(1^{2} + 5)/4 = 1(1 + 5)/4 = 1(6)/4 = 3/2

a_{2} = 2(2^{2} + 5)/4 = (4 + 5)/2 = 9/2

a_{3} = 3(3^{2} + 5)/4 = 3(9 + 5)/4 = 3(14)/4 = 21/2

a_{4} = 4(4^{2} + 5)/4 = (16 + 5) = 21

a_{5} = 5(5^{2} + 5)/4 = 5(25 + 5)/4 = 5(30)/4 = 75/2

**Ans:** The first five terms are 3/2, 9/2, 21/2, 21 and 75/2.

#### Example – 11:

- A sequence defined by a
_{n}= n^{3}– 6n^{2}+ 11n – 6, n ∈ N. Show that the first three terms of sequence are zero and all other terms are positive. **Solution:**

Given sequence is a_{n} = n^{3} – 6n^{2} + 11n – 6, n ∈ N

By substituting n = 1, 2, 3, 4 we get first four terms

a_{1} = 1^{3} – 6 x 1^{2} + 11 x 1 – 6 = 1 – 6 + 11 – 6 = 0

a_{2} = 2^{3} – 6 x 2^{2} + 11 x 2 – 6 = 8 – 24 + 22 – 6 = 0

a_{3} = 3^{3} – 6 x 3^{2} + 11 x 3 – 6 = 27 – 54 + 33 – 6 = 0

a_{4} = 4^{3} – 6 x 4^{2} + 11 x 4 – 6 = 64 – 96 + 44 – 6 = 6

a_{5} = 5^{3} – 6 x 5^{2} + 11 x 5 – 6 = 125 – 150 + 55 – 6 = 24

a_{5} = 6^{3} – 6 x 6^{2} + 11 x 6 – 6 = 216 – 216 + 66 – 6 = 60

The term is increasing fast w.r.t negative term

hence further numbers of sequence should be positive.

**Ans:** Hence it is proved that the first three terms of the sequence

are zero and all other terms are positive.

#### Example – 12:

- Find the first four terms of sequence whose first term is 1 and value of (n + 1)
^{th}term is obtained by subtracting n from n^{th}term. **Solution:**

Given a_{1} = 1 and a_{n+ 1} = a_{n }– n

By substituting n = 2, 3, 4 we get next three terms

a_{1 + 1} = a_{2 } = a_{1 }– 1 = 1 – 1 = 0

a_{2 + 1} = a_{3 } = a_{2 }– 2 = 0 – 2 = – 2

a_{3 + 1} = a_{4 } = a_{3 }– 3 = – 2 – 3 = – 5

**Ans:** The first four terms of the sequence are 1, 0, – 2, – 5.

#### Example – 13:

- Find the first four terms of sequence defined a
_{1}= 3 and a_{n}= 3a_{n- 1}+ 2 for all n > 1 **Solution:**

Given a_{1} = 3 and a_{n} = 3a_{n- 1} + 2

By substituting n = 2, 3, 4 we get next three terms

a_{2} = 3a_{2- 1} + 2 = 3a_{1} + 2 = 3 x 3 + 2 = 9 + 2 = 11

a_{3} = 3a_{3- 1} + 2 = 3a_{2} + 2 = 3 x 11 + 2 = 33 + 2 = 35

a_{4} = 3a_{4- 1} + 2 = 3a_{3} + 2 = 3 x 35 + 2 = 105 + 2 = 107

**Ans:** The first four terms of the sequence are 3, 11, 35, 107

#### Example – 14:

- Find the first five terms of sequence defined a
_{1}= 1 and a_{n}= a_{n- 1}+ 2 for all n ≥ 2 **Solution:**

Given a_{1} = 1 and a_{n} = a_{n- 1} + 2

By substituting n = 2, 3, 4, 5 we get next four terms

a_{2} = a_{2- 1} + 2 = a_{1} + 2 = 1 + 2 = 3

a_{3} = a_{3 – 1} + 2 = a_{2} + 2 = 3 + 2 = 5

a_{4} = a_{4 – 1} + 2 = a_{3} + 2 = 5 + 2 = 7

a_{5} = a_{5 – 1} + 2 = a_{4} + 2 = 7 + 2 = 9

**Ans:** The first five terms of the sequence are 1, 3, 5, 7, 9

#### Example – 15:

- Find the first five terms of sequence defined a
_{1}= a_{2}= 1 and a_{n}= a_{n- 1}+ a_{n- 2 }for all n > 2 **Solution:**

Given a_{1} = a_{2} =1 and a_{n} = a_{n- 1} + a_{n- 2}

By substituting n = 3, 4, 5 we get next three terms

a_{3} = a_{3- 1} + a_{3- 2 }= a_{2} + a_{1} = 1 + 1 = 2

a_{4} = a_{4- 1} + a_{4- 2 }= a_{3} + a_{2} = 2 + 1 = 3

a_{5} = a_{5- 1} + a_{5- 2 }= a_{4} + a_{3} = 3 + 2 = 5

**Ans:** The first five terms of the sequence are 1, 1, 2, 3, 5

#### Example – 21:

- Find the first five terms of sequence defined a
_{1}= a_{2}= 2 and a_{n}= a_{n- 1}– 1_{ }for all n > 2 **Solution:**

Given a_{1} = a_{2} = 2 and a_{n} = a_{n- 1} – 1

By substituting n = 3, 4, 5 we get next three terms

a_{3} = a_{3- 1} – 1= a_{2} – 1 = 2 – 1 = 1

a_{4} = a_{4- 1} – 1= a_{3} – 1 = 1 – 1 = 0

a_{5} = a_{5- 1} – 1= a_{4} – 1 = 0 – 1 = – 1

**Ans:** The first five terms of the sequence are 2, 2, 1, 0, – 1

#### Example – 16:

- Find the first five terms of sequence defined a
_{1}= – 1 and a_{n}= a_{n- 1}/n and n ≥ 2 **Solution:**

Given a_{1} = -1 and a_{n} = a_{n- 1} /n and n ≥ 2

By substituting n = 2, 3, 4, and 5 we get next four terms

a_{2} = a_{2- 1} /2 = a_{1} /2 = -1/2

a_{3} = a_{3- 1} /3 = a_{2} /3 = (-1/2)/3 = – 1/6

a_{4} = a_{4- 1} /4 = a_{3} /4 = (-1/6)/4 = – 1/24

a_{5} = a_{5- 1} /5 = a_{4} /5 = (-1/24)/5 = – 1/120

**Ans:** The first five terms of the sequence are -1, -1/2, -1/6, -1/24, and -1/120.

#### Example – 17:

- The Fibonacci sequence is defined by a
_{1}= a_{2}= 1 and a_{n}= a_{n- 1}+ a_{n- 2 }for all n > 2. Find the ratio a_{n + 1}/a_{n}for n = 1, 2, 3, 4, 5. **Solution:**

Given a_{1} = a_{2} =1 and a_{n} = a_{n- 1} + a_{n- 2}

By substituting n = 3, 4, 5 we get next three terms

a_{3} = a_{3- 1} + a_{3- 2 }= a_{2} + a_{1} = 1 + 1 = 2

a_{4} = a_{4- 1} + a_{4- 2 }= a_{3} + a_{2} = 2 + 1 = 3

a_{5} = a_{5- 1} + a_{5- 2 }= a_{4} + a_{3} = 3 + 2 = 5

a_{6} = a_{6- 1} + a_{6- 2 }= a_{5} + a_{4} = 5 + 3 = 8

Ratio a_{n + 1}/a_{n} for n = 1

a_{1 + 1}/a_{1} = a_{2}/a_{1} = 1/1 = 1

Ratio a_{n + 1}/a_{n} for n = 2

a_{2 + 1}/a_{2} = a_{3}/a_{2} = 2/1 = 2

Ratio a_{n + 1}/a_{n} for n = 3

a_{3 + 1}/a_{3} = a_{4}/a_{3} = 3/2 = 1.5

Ratio a_{n + 1}/a_{n} for n = 4

a_{4 + 1}/a_{4} = a_{5}/a_{4} = 5/3 = 1.667

Ratio a_{n + 1}/a_{n} for n = 5

a_{5 + 1}/a_{5} = a_{6}/a_{5} = 8/5 = 1.60

**Ans:** The ratios a_{n + 1}/a_{n} for n = 1, 2, 3, 4, 5 are 1, 2, 1.5, 1.667, 1.6

### Type – IB: To Find Indicated Term of Sequence

#### Example – 18:

- What is the 20
^{th}term of the sequence defined by a_{n}= (n – 1) (2 – n) (3 + n) ? **Solution:**

Given a_{n} = (n – 1) (2 – n) (3 + n)

By substituting n = 20 we get the 20^{th} term

a_{20} = (20 – 1) (2 – 20) (3 + 20) = (19)(-18)(23) = – 7866

**Ans:** The 20^{th} term is – 7866

#### Example – 19:

- What is the 17
^{th}and 24^{th}term of the sequence defined by a_{n}= (4n – 3) ? **Solution:**

Given a_{n} = (4n – 3)

By substituting n = 17 we get the 17^{th} term

a_{17} = 4 x 17 – 3 = 68 – 3 = 65

By substituting n = 24 we get the 24^{th} term

a_{24} = 4 x 24 – 3 = 96 – 3 = 93

**Ans:** The 17^{th} and 24^{th }terms are 65 and 93 respectively.

#### Example – 20:

- What is the 7
^{th}term of the sequence defined by a_{n}= n^{2}/2^{n}. **Solution:**

Given a_{n} = n^{2}/2^{n}

By substituting n = 7 we get the 7^{th} term

a_{7} = 7^{2}/2^{7} = 49/128

**Ans:** The 7^{th} term is 49/128.

#### Example – 21:

- What is the 9
^{th}term of the sequence defined by a_{n}= (-1)^{n – 1 }n^{3}. **Solution:**

Given a_{n} = (-1)^{n – 1 }n^{3}

By substituting n = 9 we get the 9^{th} term

a_{9} = (-1)^{9- 1 }n^{3} = (-1)^{8 }9^{3}. = (1)(729) = 729

**Ans:** The 9^{th} term is 729.

#### Example – 22:

- What is the 20
^{th}term of the sequence defined by a_{n}= n(n – 2)/(n + 3) **Solution:**

Given a_{n} = n(n – 2)/(n + 3)

By substituting n = 20 we get the 20^{th} term

a_{9} = 20(20 – 2)/(20 + 3) = 20(18)/23 = 360/23

**Ans:** The 20^{th} term is 360/23.

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

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