Question 1: If the n^{th} term a_n of a sequence is given by a_n = n^2-n+1 , write down the first five terms.

Answer:

Given sequence is a_n = n^2-n+1

Let the first five terms be a_1, a_2, a_3, a_4, a_5

a_1 = (1)^2 - 1 + 1 = 1

a_2 = (2)^2 - 2 + 1 = 3

a_3 = (3)^2 - 3 + 1 = 7

a_4 = (4)^2 - 4 + 1 = 13

a_5 = (5)^2 - 5 + 1 = 21

Therefore the first five terms are 1, 3, 7, 13, and 21

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Question 2: A sequence is defined by a_n = n^3 - 6n^2 + 11n -6, n \in N . Show that the first three terms of the sequence are zero and all other terms are positive.

Answer:

The given sequence is a_n = n^3 - 6n^2 + 11n -6, n \in N

Let the first five terms be a_1, a_2, a_3

a_1 = (1)^3 - 6(1)^2 + 11(1) - 6  = 0

a_1 = (2)^3 - 6(2)^2 + 11(2) - 6  = 0

a_1 = (3)^3 - 6(3)^2 + 11(3) - 6  = 0

Therefore the first three terms are 0

And a_n = n^3 - 6n^2 + 11n -6 = (n-2)^3 - (n-2) 

Therefore a_n will be positive for n \geq 4

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Question 3: Find the first four terms of the sequence defined by a_1 =3 and a_n = 3a_{n-1} +2 for all.

Answer:

Given a_1 =3 and a_n = 3a_{n-1} +2

a_2 = 3 a_{2-1}+ 2 = 3a_1 + 2 = 3(3) + 2 = 11

a_3 = 3 a_{3-1}+ 2 = 3a_2 + 2 = 3(11) + 2 = 35

a_4 = 3 a_{4-1}+ 2 = 3a_3 + 2 = 3(35) + 2 = 107

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Question 4: Write the first five terms in each of the following sequences:

i) a_1 = 1, a_n = a_{n-1} + 2, n > 1

ii) a_1 = 1 = a_2, a_n = a_{n-1}+ a_{n-2}, n > 2

iii)  a_1=a_2= 2, a_n = a_{n-1}-1, n > 2

Answer:

i)      a_1 = 1, a_n = a_{n-1} + 2, n > 1

a_1 = 1

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

Therefore the first five terms of the given sequence are 1, 3, 5, 7 and 9 .

ii)      a_1 = 1 = a_2, a_n = a_{n-1}+ a_{n-2}, n > 2

a_1 = 1

a_2 = 1

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

Therefore the first five terms of the given sequence are 1, 1, 2, 3 and 5 .

iii)    a_1=a_2= 2, a_n = a_{n-1}-1, n > 2

a_1 = 2

a_2 = 2

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

Therefore the first five terms of the given sequence are 2, 2, 1, 0 and -1 .

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Question 5: The Fibonacci sequence is defined by a_1 = 1 = a_2, a_n = a_{n-1} + a_{n-2} for n > 2 . Find \frac{a_{n + 1}}{a_n} for n = 1, 2, 3, 4, 5 .

Answer:

Given: a_1= 1,   \ \ a_2 = 1

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

Therefore for \frac{a_{n + 1}}{a_n}

For n = 1: \ \ \frac{a_{1 + 1}}{a_1} = \frac{a_2}{a_1} = \frac{1}{1} = 1

For n = 2: \ \ \frac{a_{2 + 1}}{a_2} = \frac{a_3}{a_2} = \frac{2}{1} = 2

For n = 3: \ \ \frac{a_{3 + 1}}{a_3} = \frac{a_4}{a_3} = \frac{3}{2}

For n = 4: \ \ \frac{a_{4 + 1}}{a_4} = \frac{a_5}{a_4} = \frac{5}{3}

For n = 5: \ \ \frac{a_{5 + 1}}{a_5} = \frac{a_6}{a_5} = \frac{8}{5}

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Question 6: Show that each of the following sequences is in A.P. Also find the common difference and write 3 more terms in each case.

i) 3, -1, -5, -9, \ldots

ii) -1, \frac{1}{4} , \frac{3}{2} , \frac{11}{4} , \ldots

iii) \sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}, \ldots

iv)  9, 7 , 5, 3 \ldots

Answer:

i)       Given sequence: 3, -1, -5, -9, \ldots

a_1 = 3, \ \ a_2 = -1, \ \ a_3 = -5, \ \ a_4 = -9

a_2 - a_1 = -1-3 = - 4

a_3 - a_2 = -5 - (-1) = - 4

a_4 - a_3 = -9 - (-5) = - 4

Therefore the common difference (d) = -4

a_5 = a_4 + d = -9 +(-4) = - 13

a_6 = a_5 + d = -13 +(-4) = -17

a_7 = a_6 + d = -17 +(-4) = -21

ii)      Given sequence: -1, \frac{1}{4} , \frac{3}{2} , \frac{11}{4} , \ldots

a_1 = -1, \ \ a_2 = \frac{1}{4} , \ \ a_3 = \frac{3}{2} , \ \ a_4 = \frac{11}{4}

a_2 - a_1 = \frac{1}{4} - (-1) =  \frac{5}{4}

a_3 - a_2 = \frac{3}{2} - ( \frac{1}{4} ) =  \frac{5}{4}

a_4 - a_3 = \frac{11}{4} - ( \frac{3}{2} ) =  \frac{5}{4}

Therefore the common difference (d) = \frac{5}{4}

a_5 = a_4 + d = \frac{11}{4} + \frac{5}{4} = 4

a_6 = a_5 + d = 4 + \frac{5}{4} = \frac{21}{4}

a_7 = a_6 + d = \frac{21}{4} + \frac{5}{4} = \frac{26}{4} = \frac{13}{2}

iii)    Given sequence: \sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}, \ldots

a_1 = \sqrt{2}, \ \ a_2 = 3\sqrt{2}, \ \ a_3 = 5\sqrt{2}, \ \ a_4 = 7\sqrt{2}

a_2 - a_1 = 3\sqrt{2}-\sqrt{2} = 2\sqrt{2}

a_3 - a_2 = 5\sqrt{2} - 3\sqrt{2} = 2\sqrt{2}

a_4 - a_3 = 7\sqrt{2} - 5\sqrt{2} = 2\sqrt{2}

Therefore the common difference (d) = 2\sqrt{2}

a_5 = a_4 + d = 7\sqrt{2} +2\sqrt{2} = 9\sqrt{2}

a_6 = a_5 + d = 9\sqrt{2} +2\sqrt{2} = 11\sqrt{2}

a_7 = a_6 + d = 11\sqrt{2} +2\sqrt{2} = 13\sqrt{2}

iv)     Given sequence: 9, 7 , 5, 3 \ldots

a_1 = 9, \ \ a_2 = 7, \ \ a_3 = 5, \ \ a_4 = 3

a_2 - a_1 = 7-9 = - 2

a_3 - a_2 = 5-7 = - 2

a_4 - a_3 = 3-5 = - 2

Therefore the common difference (d) = -2

a_5 = a_4 + d = 3 +(-2) = 1

a_6 = a_5 + d = 1 +(-2) = -1

a_7 = a_6 + d = -1 +(-2) = -3

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Question 7: The n^{th} terms of a sequence is given by a_n = 2n+ 7 . Show that it is an A.P. Also, find its 7^{th} terms.

Answer:

Given sequence a_n = 2n+ 7

a_1 = 2 ( 1) + 7 = 9

a_2 = 2(2) + 7 = 11

a_3 = 2(3) + 7 = 13

a_2 - a_1 = a_3 - a_2 = 2

Therefore the common difference (d) = 2 . Therefore the given sequence is an A.P.

a_7 = 2(7) + 7 = 21

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Question 8: The n^{th} terms of a sequence is given by a_n = 2n^2 +n + 1 . Show that it is not an A.P.

Answer:

Given sequence a_n = 2n^2 +n + 1

a_1 = 2(1)^2 + 1 + 1 = 4

a_2 = 2(2)^2 + 2 + 1 = 11

a_3 = 2(3)^2 + 3 + 1 = 21

Therefore a_3-a_2 \neq a_2-a_1

Hence the sequence is not an A.P.