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Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progression Exercise 9.1

  • Last Updated : 04 Dec, 2020

Problem 1: Write the first terms of each of the following sequences whose nth term are:

(i) an = 3n + 2

Solution:

Given:

an = 3n + 2

By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a1 = (3 × 1) + 2 = 3 + 2 = 5

a2 = (3 × 2) + 2 = 6 + 2 = 8

a3 = (3 × 3) + 2 = 9 + 2 = 11

a4 = (3 × 4) + 2 = 12 + 2 = 14

a5 = (3 × 5) + 2 = 15 + 2 = 17

∴ The first five terms are 5, 8, 11, 14, 17

(ii) an = (n – 2)/3

Solution: 

Given:

By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

 an = (n – 2)/3

a1 = (1-2)/3 = -1/3

a2 = (2 – 2)/3 = 0

a3 = (3 – 2)/3 = 1/3

a4 = (4 – 2)/3 = 2/3

a5 = (5 – 2)/3 = 3/3 =1 

∴ The first five terms are -1/3, 0, 1/3, 2/3, 1

(iii) an = 3n

Solution:

Given: 

By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a1 = 31 = 3;

a2 = 32 = 9;

a3 = 33  = 27 ;

a4 = 34 = 81;

a5 = 35 = 243.

∴ The first five terms are 3, 9, 27, 81, 243.

(iv) an = (3n – 2)/ 5

Solution:

Given:

By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a1 = (3 * 1 – 2)/5 = 1/5

a2 = (3 * 2 – 2)/5 = 4/5

a3 = (3 * 3 – 2)/5  = 7/5

a4 = (3* 4 – 2)/5= 10/5 =2

a5 = (3 * 5 – 2)/5 =13/5 

∴ The first five terms are 1/5, 4/5, 7/5, 2, 13/5

(v) an = (-1)n2n

Solution:

Given:

By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a1 = (-1)1 . 2 = -2

a2 = (-1)2 . 22  = 4

a3 = (-1)3 . 23   = -8

a4 = (-1)4 . 2 = 16

a5 = (-1)5 . 25  = -32

∴ The first five terms are -2, 4, -8, 16, -32

(vi) an = n(n – 2)/2

Solution:

Given:

By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a1 = (1.( 1 – 2))/2 = -1/2

a2 = (2.(2 – 2))/2 = 0

a3 = (3.(3 – 2))/2 = 3/2

a4 = (4.(4 – 2))/2 = 4

a5 = (5.(5 – 2))/2 =15/2

∴ The first five terms are -1/2, 0, 3/2, 4, 15/2

(vii) an = n2 – n + 1

Solution: 

Given:

By putting n = 1, 2, 3, 4, 5  we get first five term of the sequence

a1 = 12 – 1 + 1 = 1

a2 = 22 – 2 + 1 = 3

a3 = 32 – 3 + 1  = 7

a4 = 42 – 4 + 1 = 13

a5 = 52 – 5 + 1 = 21

∴ The first five terms are  1, 3, 7, 13, 21

(viii) an = 2n2 – 3n + 1

Solution:

Given:

By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence

a1 = 2 .12 – 3.1 + 1 = 0

a2 = 2. 22 – 3.2 + 1 = 3

a3 = 2.32 – 3.3 + 1 = 10

a4 = 2.42 – 3.4 + 1 = 21

a5 = 2.52  – 3.5 + 1 = 36

∴ The first five terms are 0, 3, 10, 21, 36

Problem 2: Find the indicated terms in each of the following sequences whose nth terms are:

(i) an = 5n – 4; a12 and a15

Solution :

Given:

an = 5n – 4

a12 = By putting n=12

a12  = 5.12 – 4 = 56

a15 = By putting n=15

a15  = 5.15 – 4 = 71

∴ The required terms a12 = 56, a15 = 71

(ii) an = (3n – 2)/(4n + 5), a7 and a8

Solution:

Given:

an = (3n – 2)/(4n + 5)

a7 = By putting n=7

     = (3.7 – 2)/(4.7 + 5)

      =19/33

a8 = By putting n = 8

       = (3.8 – 2)/(4.8 + 5)

       = 2 2/37

∴ The required terms a7 = 19/33, a8 = 22/37

 (iii) an = n(n – 1)(n – 2); a5 and a8

Solution: 

Given:

an = n(n – 1)(n – 2)

By putting n=5

a5 = 5(5 – 1).(5 – 2) = 5.4.3 = 60

By putting n=8

a8 = 8.(8 – 1).(8 – 2) = (8.7.6) = 336

∴ The required terms a5= 60, a8 = 336

(iv) an = (n – 1)(2 – n)(3 + n); a1, a2, a

Solution:

Given:

an = (n – 1)(2 – n)(3 + n)

By putting n = 1

a1 = (1 – 1)(2 – 1)(3 + 1) = 0.1.4 = 0

By putting n = 2

a2 = (2 – 1)(2 – 2)(3 + 2) = 1.0.5 = 0

By putting n = 3

a3 = (3 – 1)(2 – 3)(3 + 3) = 2.-1.6 = -12

∴ The required terms a1= 0, a2 = 0 and a3 = -12

(v) an = (-1)n.n; a3, a5, a8

Solution:

Given:

 an = (-1)n.n

By putting n=3

a3 = (-1)3.3 = -1.3 = -3

By putting n=5

a5 = (-1)5.5 = -1.5 = -5

By putting n=8

a8 = (-1)8.8 = 1.8 = 8

∴ The required terms a3= -3, a5 = -5 and a8 = 8

Problem 3: Find the next five terms of each of the following sequence given by:

(i) a1 = 1, an = an – 1 + 2, n 2

Solution:

By putting n = 2, 3, 4, 5, 6 we get next five term of the sequence

a2 = a1 + 2 = 1 + 2 = 3

a3 = a2 + 2 = 3 + 2 = 5

a4 = a3 + 2 = 5 + 2 = 7

a5 = a4 + 2 = 7 + 2 = 9

a6 = a5 + 2 = 9 + 2 = 11

∴ The next five terms are 3, 5, 7, 9, 11

(ii) a1 = a2 = 2, an = an – 1 – 3, n > 2

Solution:

By putting n = 3, 4, 5, 6, 7 we get next five term of the sequence

a3 = a2 – 3 = 2 – 3 = -1

a4 = a3 – 3 = -1 – 3 = -4

a5 = a4 – 3 = -4 – 3 = -7

a6 = a5 – 3 = -7 – 3 = -10

a7 = a6 – 3 = -10 – 3 = -13

∴ The next five terms are -1, -4, -7, -10, -13

(iii) a1 = -1, an = (an – 1)/(n), n >= 2

Solution:

By putting n = 2, 3, 4, 5, 6  we get next five term of the sequence

a2 = a1 /2 = -1/2

a3 = a2 /3 = (-1/2)/3 = -1/6 

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

a5 = a4/5 = (-1/24)/5 = -1/120

a6 = a5/6 =(-1/120)/6 = -1/720

∴ The next five terms are -1/2, -1/6, -1/24, -1/120, -1/720

(iv) a1 = 4, an = 4an – 1 + 3, n > 1

Solution:

By putting n = 2, 3, 4, 5, 6 we get next five term of the sequence

a2 = 4.a1  + 3 = 4.4 + 3 = 19

a3 = 4.a + 3 = 4.19 + 3 = 79

a4 = 4.a + 3 = 4.79 + 3 = 319

a5 = 4.a4 + 3 = 4.319 + 3 = 1279

a6 = 4.a5 + 3 = 5.1279 + 3 = 5119

∴ The next five terms are 19, 79, 319, 1279, 5119


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