# Sequence

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• 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 a1, a2, a3, ….., aor t1, t2, t3, ….., tn
• 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 an = n/(n2 + 1)
• Solution:

Given an = n/(n2 + 1)

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

a1 = 1/(12 + 1) = 1/2

a2 = 2/(22 + 1) = 2/5

a3 = 3/(32 + 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 an = n2 – n + 1
• Solution:

Given an = n2 – n + 1

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

a1 = 12 – 1 + 1 = 1 – 1 + 1 = 1

a2 = 22 – 2 + 1 = 4 – 2 + 1 = 3

a3 = 32 – 3 + 1 = 9 – 3 + 1 = 7

a4 = 42 – 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 an = 2n + 5
• Solution:

Given an = 2n + 5

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

a1 = 2(1) + 5 = 2 + 5 = 7

a2 = 2(2) + 5 = 4 + 5 = 9

a3 = 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 an = (n – 3)/4
• Solution:

Given sequence is an = (n – 3)/4

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

a1 = (1 – 3)/4 = -2/4 = -1/2

a2 = (2 – 3)/4 = -1/4

a3 = (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 an = n(n + 2)
• Solution:

Given an = n(n + 2)

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

a1 = 1(1+2) = 1(3) = 3

a2 = 2(2+2) = 2(8) = 8

a3 = 3(3+2) = 3(5) = 15

a4 = 4(4+2) = 4(6) =  24

a5 = 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 an = n/(n + 1)
• Solution:

Given an = n/(n + 1)

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

a1 = 1/(1+1) = 1/2

a2 = 2/(2+1) = 2/3

a3 = 3/(3+1) = 3/4

a4 = 4/(4+1) = 4/5

a5 = 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 an = 2n
• Solution:

Given an = 2n

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

a1 = 21 = 2

a2 = 22 = 4

a3 = 23 = 8

a4 = 24 = 16

a5 = 25 = 32

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

#### Example – 08:

• Give first five terms of the sequence defined by an = (2n – 3)/6
• Solution:

Given an = (2n – 3)/6

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

a1 = (2 x 1 – 3)/6 = (2 – 3)/6 = -1/6

a2 = (2 x 2 – 3)/6 = (4 – 3)/6 = 1/6

a3 = (2 x 3 – 3)/6 = (6 – 3)/6 = 3/6 = 1/2

a4 = (2 x 4 – 3)/6 = (8 – 3)/6 = 5/6

a5 = (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 an = (-1)n – 1 5n + 1
• Solution:

Given an = (-1)n – 1 5n + 1

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

a1 = (-1)1 – 1 51 + 1 = (-1)52 = 1 x 25 = 25

a2 = (-1)2 – 1 52 + 1 = (-1)53 = – 1 x 125 = – 125

a3 = (-1)3 – 1 53 + 1 = (-1)54 = 1 x 625 = 625

a4 = (-1)4 – 1 54 + 1 = (-1)55 = -1 x 3125 = – 3125

a5 = (-1)5 – 1 55 + 1 = (-1)56 = 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 an = n(n2 + 5)/4
• Solution:

Given an = n(n2 + 5)/4

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

a1 = 1(12 + 5)/4 = 1(1 + 5)/4 = 1(6)/4 = 3/2

a2 = 2(22 + 5)/4 = (4 + 5)/2 = 9/2

a3 = 3(32 + 5)/4 = 3(9 + 5)/4 =  3(14)/4 = 21/2

a4 = 4(42 + 5)/4 = (16 + 5) =  21

a5 = 5(52 + 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 an = n3 – 6n2 + 11n – 6, n ∈ N. Show that the first three terms of sequence are zero and all other terms are positive.
• Solution:

Given sequence is an =  n3 – 6n2 + 11n – 6, n ∈ N

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

a1 =  13 – 6 x 12 + 11 x 1 – 6 = 1 – 6 + 11 – 6 = 0

a2 =  23 – 6 x 22 + 11 x 2 – 6 = 8 – 24 + 22 – 6 = 0

a3 =  33 – 6 x 32 + 11 x 3 – 6 = 27 – 54 + 33 – 6 = 0

a4 =  43 – 6 x 42 + 11 x 4 – 6 = 64 – 96 + 44 – 6 = 6

a5 =  53 – 6 x 52 + 11 x 5 – 6 = 125 – 150 + 55 – 6 = 24

a5 =  63 – 6 x 62 + 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 nth term.
• Solution:

Given a1 = 1 and an+ 1 = a– n

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

a1 + 1 = a = a– 1 = 1 – 1 = 0

a2 + 1 = a = a– 2 = 0 – 2 = – 2

a3 + 1 = a = a– 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 a1 = 3 and an = 3an- 1 + 2 for all n > 1
• Solution:

Given a1 = 3 and an = 3an- 1 + 2

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

a2 = 3a2- 1 + 2 = 3a1 + 2 = 3 x 3 + 2 = 9 + 2 = 11

a3 = 3a3- 1 + 2 = 3a2 + 2 = 3 x 11 + 2 = 33 + 2 = 35

a4 = 3a4- 1 + 2 = 3a3 + 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 a1 = 1 and an = an- 1 + 2 for all n ≥ 2
• Solution:

Given a1 = 1 and an = an- 1 + 2

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

a2 = a2- 1 + 2 = a1 + 2 = 1 + 2 = 3

a3 = a3 – 1 + 2 = a2 + 2 = 3 + 2 = 5

a4 = a4 – 1 + 2 = a3 + 2 = 5 + 2 = 7

a5 = a5 – 1 + 2 = a4 + 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 a1 =  a2 = 1 and an = an- 1 + an- 2  for all n > 2
• Solution:

Given a1 =  a2 =1 and an = an- 1 + an- 2

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

a3 = a3- 1 + a3- 2 = a2 + a1 =  1 + 1 = 2

a4 = a4- 1 + a4- 2 = a3 + a2 =  2 + 1 = 3

a5 = a5- 1 + a5- 2 = a4 + a3 =  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 a1 =  a2 = 2 and an = an- 1 – 1  for all n > 2
• Solution:

Given a1 =  a2 = 2 and an = an- 1  – 1

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

a3 = a3- 1  – 1= a2 – 1 =  2 – 1 = 1

a4 = a4- 1  – 1= a3 – 1 =  1 – 1 = 0

a5 = a5- 1  – 1= a4 – 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 a1 =  – 1 and an = an- 1 /n and n ≥ 2
• Solution:

Given a1 =  -1 and an = an- 1 /n and n ≥ 2

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

a2 = a2- 1 /2 = a1 /2 = -1/2

a3 = a3- 1 /3 = a2 /3 = (-1/2)/3 = – 1/6

a4 = a4- 1 /4 = a3 /4 = (-1/6)/4 = – 1/24

a5 = a5- 1 /5 = a4 /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 a1 =  a2 = 1 and an = an- 1 + an- 2  for all n > 2. Find the ratio an + 1/an for n = 1, 2, 3, 4, 5.
• Solution:

Given a1 =  a2 =1 and an = an- 1 + an- 2

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

a3 = a3- 1 + a3- 2 = a2 + a1 =  1 + 1 = 2

a4 = a4- 1 + a4- 2 = a3 + a2 =  2 + 1 = 3

a5 = a5- 1 + a5- 2 = a4 + a3 =  3 + 2 = 5

a6 = a6- 1 + a6- 2 = a5 + a4 =  5 + 3 = 8

Ratio an + 1/an for n = 1

a1 + 1/a1 = a2/a1 = 1/1 = 1

Ratio an + 1/an for n = 2

a2 + 1/a2 = a3/a2 = 2/1 = 2

Ratio an + 1/an for n = 3

a3 + 1/a3 = a4/a3 = 3/2 = 1.5

Ratio an + 1/an for n = 4

a4 + 1/a4 = a5/a4 = 5/3 = 1.667

Ratio an + 1/an for n = 5

a5 + 1/a5 = a6/a5 = 8/5 = 1.60

Ans: The ratios an + 1/an 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 20th term of the sequence defined by an = (n – 1) (2 – n) (3 + n) ?
• Solution:

Given an(n – 1) (2 – n) (3 + n)

By substituting n = 20 we get the 20th term

a20(20 – 1) (2 – 20) (3 + 20) = (19)(-18)(23) = – 7866

Ans: The  20th term is – 7866

#### Example – 19:

• What is the 17th and 24th term of the sequence defined by an = (4n – 3) ?
• Solution:

Given an(4n – 3)

By substituting n = 17 we get the 17th term

a17 =  4 x 17 – 3 = 68 – 3 = 65

By substituting n = 24 we get the 24th term

a24 =  4 x 24 – 3 = 96 – 3 = 93

Ans: The  17th and 24th terms are 65 and 93 respectively.

#### Example – 20:

• What is the 7th term of the sequence defined by an = n2/2n.
• Solution:

Given an = n2/2n

By substituting n = 7 we get the 7th term

a772/27 = 49/128

Ans: The  7th term is 49/128.

#### Example – 21:

• What is the 9th term of the sequence defined by an = (-1)n – 1 n3.
• Solution:

Given an = (-1)n – 1 n3

By substituting n = 9 we get the 9th term

a9 (-1)9- 1 n3 = (-1)93. = (1)(729) = 729

Ans: The  9th term is 729.

#### Example – 22:

• What is the 20th term of the sequence defined by an = n(n – 2)/(n + 3)
• Solution:

Given an = n(n – 2)/(n + 3)

By substituting n = 20 we get the 20th term

a920(20 – 2)/(20 + 3) = 20(18)/23 = 360/23

Ans: The  20th term is 360/23.

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