Sequence

A sequence is a function whose domain is the set N of natural numbers. The sequence is generally denoted by a and its image is denoted by a(n) or t(n). The terms of the sequence are represented by a₁, a₂, a₃, ….., a_n or t₁, t₂, t₃, ….., t_n

• A sequence whose range is a subset of R (set of real numbers) is called a real sequence.
• The 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 commodities occur in a sequence.
• Sequences have important applications in many human activities.
• A sequence containing a finite number of terms is called a finite sequence. 
• A sequence is called infinite, if it is not a finite sequence. 

To Find First Three/ Five Terms of Sequence:

Example – 01:

Give the first three terms of the sequence defined by a_n = n/(n² + 1)

Solution:

Given a_n = 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 the first four terms of the sequence defined by a_n = n² – n + 1

Solution:

Given a_n = 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 the 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

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 the 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

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 the 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

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 the 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

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 the first five terms of the sequence defined by a_n = 2ⁿ

Solution:

Given a_n = 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 the 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

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 the 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

a₁ = (-1)^{1 – 1} 5^{1 + 1} = (-1)⁰ 5² = 1 x 25 = 25

a₂ = (-1)^{2 – 1} 5^{2 + 1} = (-1)¹ 5³ = – 1 x 125 = – 125

a₃ = (-1)^{3 – 1} 5^{3 + 1} = (-1)² 5⁴ = 1 x 625 = 625

a₄ = (-1)^{4 – 1} 5^{4 + 1} = (-1)³ 5⁵ = -1 x 3125 = – 3125

a₅ = (-1)^{5 – 1} 5^{5 + 1} = (-1)⁴ 5⁶ = 1 x 15625 = 15625

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

Example – 10:

Give the first five terms of the sequence defined by a_n = n(n² + 5)/4

Solution:

Given a_n = 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 a_n = 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 a_n =  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 a sequence whose first term is 1 and the value of (n + 1)^th term is obtained by subtracting n from n^th term.

Solution:

Given a₁ = 1 and a_{n+ 1} = a_n – n

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

a_{1 + 1} = a₂  = a₁ – 1 = 1 – 1 = 0

a_{2 + 1} = a₃  = a₂ – 2 = 0 – 2 = – 2

a_{3 + 1} = a₄  = a₃ – 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 a_n = 3a_{n- 1} + 2 for all n > 1

Solution:

Given a₁ = 3 and a_n = 3a_{n- 1} + 2

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

a₂ = 3a_{2- 1} + 2 = 3a₁ + 2 = 3 x 3 + 2 = 9 + 2 = 11

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

a₄ = 3a_{4- 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 a_n = a_{n- 1} + 2 for all n ≥ 2

Solution:

Given a₁ = 1 and a_n = a_{n- 1} + 2

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

a₂ = a_{2- 1} + 2 = a₁ + 2 = 1 + 2 = 3

a₃ = a_{3 – 1} + 2 = a₂ + 2 = 3 + 2 = 5

a₄ = a_{4 – 1} + 2 = a₃ + 2 = 5 + 2 = 7

a₅ = a_{5 – 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 a_n = a_{n- 1} + a_{n- 2}  for all n > 2

Solution:

Given a₁ =  a₂ =1 and a_n = a_{n- 1} + a_{n- 2}

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

a₃ = a_{3- 1} + a_{3- 2} = a₂ + a₁ =  1 + 1 = 2

a₄ = a_{4- 1} + a_{4- 2} = a₃ + a₂ =  2 + 1 = 3

a₅ = a_{5- 1} + a_{5- 2} = a₄ + a₃ =  3 + 2 = 5

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

Example – 16:

Find the first five terms of sequence defined a₁ =  a₂ = 2 and a_n = a_{n- 1} – 1 for all n > 2

Solution:

Given a₁ =  a₂ = 2 and a_n = a_{n- 1}  – 1

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

a₃ = a_{3- 1}  – 1= a₂ – 1 =  2 – 1 = 1

a₄ = a_{4- 1}  – 1= a₃ – 1 =  1 – 1 = 0

a₅ = a_{5- 1}  – 1= a₄ – 1 =  0 – 1 = – 1

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

Example – 17:

Find the first five terms of sequence defined a₁ =  – 1 and a_n = a_{n- 1} /n and n ≥ 2

Solution:

Given a₁ =  -1 and a_n = a_{n- 1} /n and n ≥ 2

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

a₂ = a_{2- 1} /2 = a₁ /2 = -1/2

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

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

a₅ = a_{5- 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 – 18:

The Fibonacci sequence is defined by a₁ =  a₂ = 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 a₁ =  a₂ =1 and a_n = a_{n- 1} + a_{n- 2}

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

a₃ = a_{3- 1} + a_{3- 2} = a₂ + a₁ =  1 + 1 = 2

a₄ = a_{4- 1} + a_{4- 2} = a₃ + a₂ =  2 + 1 = 3

a₅ = a_{5- 1} + a_{5- 2} = a₄ + a₃ =  3 + 2 = 5

a₆ = a_{6- 1} + a_{6- 2} = a₅ + a₄ =  5 + 3 = 8

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

a_{1 + 1}/a₁ = a₂/a₁ = 1/1 = 1

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

a_{2 + 1}/a₂ = a₃/a₂ = 2/1 = 2

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

a_{3 + 1}/a₃ = a₄/a₃ = 3/2 = 1.5

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

a_{4 + 1}/a₄ = a₅/a₄ = 5/3 = 1.667

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

a_{5 + 1}/a₅ = a₆/a₅ = 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

To Find Indicated Term of Sequence:

Example – 19:

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

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

Ans: The  20^th term is – 7866

Example – 20:

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

a₁₇ =  4 x 17 – 3 = 68 – 3 = 65

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

a₂₄ =  4 x 24 – 3 = 96 – 3 = 93

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

Example – 21:

What is the 7^th term of the sequence defined by a_n = n²/2ⁿ.

Solution:

Given a_n = n²/2ⁿ

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

a₇ =  7²/2⁷ = 49/128

Ans: The  7^th term is 49/128.

Example – 22:

What is the 9^th term of the sequence defined by a_n = (-1)^{n – 1} n³.

Solution:

Given a_n = (-1)^{n – 1} n³

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

a₉ =  (-1)^{9- 1} n³ = (-1)⁸ 9³. = (1)(729) = 729

Ans: The  9^th term is 729.

Example – 23:

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

a₉ =  20(20 – 2)/(20 + 3) = 20(18)/23 = 360/23

Ans: The  20^th term is 360/23.