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# RD Chapter 19- Arithmetic Progressions Ex-19.1 Interview Questions Answers

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Question 1 :

If the nth term of a sequence is given by an =n2 – n+1, write down its first five terms.

Given:

an = n2 –n+1

By using the values n= 1, 2, 3, 4, 5 we can find the first five terms.

When n = 1:

a1 =(1)2 – 1 + 1

= 1 – 1 + 1

= 1

When n = 2:

a2 =(2)2 – 2 + 1

= 4 – 2 + 1

= 3

When n = 3:

a3 =(3)2 – 3 + 1

= 9 – 3 + 1

= 7

When n = 4:

a4 =(4)2 – 4 + 1

= 16 – 4 + 1

= 13

When n = 5:

a5 =(5)2 – 5 + 1

= 25 – 5 + 1

= 21

First five termsof the sequence are 1, 3, 7, 13, 21.

Question 2 :

A sequence is defined by an = n3 – 6n2 +11n – 6, n  N. Show that the first threeterms of the sequence are zero and all other terms are positive.

Given:

an = n3 –6n2 + 11n – 6, n  N

By using the values n= 1, 2, 3 we can find the first three terms.

When n = 1:

a1 =(1)3 – 6(1)2 + 11(1) – 6

= 1 – 6 + 11 – 6

= 12 – 12

= 0

When n = 2:

a2 =(2)3 – 6(2)2 + 11(2) – 6

= 8 – 6(4) + 22 – 6

= 8 – 24 + 22 – 6

= 30 – 30

= 0

When n = 3:

a3 =(3)3 – 6(3)2 + 11(3) – 6

= 27 – 6(9) + 33 – 6

= 27 – 54 + 33 – 6

= 60 – 60

= 0

This shows that thefirst three terms of the sequence is zero.

Now, let’s check forwhen n = n:

an = n3 –6n2 + 11n – 6

= n3 –6n2 + 11n – 6 – n + n – 2 + 2

= n3 –6n2 + 12n – 8 – n + 2

= (n)3 –3×2n(n – 2) – (2)3 – n + 2

By using the formula,{(a – b)3 = (a)3 – (b)3 – 3ab(a– b)}

an =(n – 2)3 – (n – 2)

Here, n – 2 willalways be positive for n > 3

an isalways positive for n > 3

Question 3 : Find the first four terms of the sequencedefined by a1 = 3 and an = 3an–1 +2, for all n > 1.

Given:

a1 = 3and an = 3an–1 + 2, for all n > 1

By using the values n= 1, 2, 3, 4 we can find the first four terms.

When n = 1:

a1 = 3

When n = 2:

a2 =3a2–1 + 2

= 3a1 +2

= 3(3) + 2

= 9 + 2

= 11

When n = 3:

a3 =3a3–1 + 2

= 3a2 +2

= 3(11) + 2

= 33 + 2

= 35

When n = 4:

a4 =3a4–1 + 2

= 3a3 +2

= 3(35) + 2

= 105 + 2

= 107

First four termsof sequence are 3, 11, 35, 107.

Question 4 :

Write the first five terms in each of the following sequences:
(i) a1 = 1, an = an–1 + 2, n> 1

(ii) a1 = 1 = a2, an = an–1 +an–2, n > 2

(iii) a1 = a2 =2, an = an–1 –1, n > 2

(i) a1 =1, an = an–1 + 2, n > 1

By using the values n= 1, 2, 3, 4, 5 we can find the first five terms.

Given:

a1 = 1

When n = 2:

a2 = a2–1 +2

= a1 +2

= 1 + 2

= 3

When n = 3:

a3 = a3–1 +2

= a2 +2

= 3 + 2

= 5

When n = 4:

a4 = a4–1 +2

= a3 +2

= 5 + 2

= 7

When n = 5:

a5 = a5–1 +2

= a4 +2

= 7 + 2

= 9

First five termsof the sequence are 1, 3, 5, 7, 9.

(ii) a1 = 1= a2, an = an–1 + an–2, n> 2

By using the values n= 1, 2, 3, 4, 5 we can find the first five terms.

Given:

a1 = 1

a2 = 1

When n = 3:

a3 = a3–1 +a3–2

= a2 +a1

= 1 + 1

= 2

When n = 4:

a4 = a4–1 +a4–2

= a3 +a2

= 2 + 1

= 3

When n = 5:

a5 = a5–1 +a5–2

= a4 +a3

= 3 + 2

= 5

First five termsof the sequence are 1, 1, 2, 3, 5.

(iii) a1 = a2 =2,an = an–1 – 1, n > 2

By using the values n= 1, 2, 3, 4, 5 we can find the first five terms.

Given:

a1 = 2

a2 = 2

When n = 3:

a3 = a3–1 –1

= a2 –1

= 2 – 1

= 1

When n = 4:

a4 = a4–1 –1

= a3 –1

= 1 – 1

= 0

When n = 5:

a5 = a5–1 –1

= a4 –1

= 0 – 1

= -1

First five termsof the sequence are 2, 2, 1, 0, -1.

Question 5 :

The Fibonacci sequence is defined by a1 = 1 = a2,an = an–1 + an–2 for n > 2.Find (an+1)/an for n = 1, 2, 3, 4, 5.

Given:

a1 = 1

a2 = 1

an = an–1 +an–2

When n = 1:

(an+1)/a=(a1+1)/a1

= a2/a1

= 1/1

= 1

a3 = a3–1 +a3–2

= a2 +a1

= 1 + 1

= 2

When n = 2:

(an+1)/a=(a2+1)/a2

= a3/a2

= 2/1

= 2

a4 = a4–1 +a4–2

= a3 +a2

= 2 + 1

= 3

When n = 3:

(an+1)/a=(a3+1)/a3

= a4/a3

= 3/2

a5 = a5–1 +a5–2

= a4 +a3

= 3 + 2

= 5

When n = 4:

(an+1)/a=(a4+1)/a4

= a5/a4

= 5/3

a6 = a6–1 +a6–2

= a5 +a4

= 5 + 3

= 8

When n = 5:

(an+1)/a=(a5+1)/a5

= a6/a5 =8/5

Value of (an+1)/an whenn = 1, 2, 3, 4, 5 are 1, 2, 3/2, 5/3, 8/5

Todays Deals  ### RD Chapter 19- Arithmetic Progressions Ex-19.1 Contributors krishan

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