Question

Write the first five terms of each of the following sequences whose nth terms are:
(a)$a_{n}=3n+2$
(b)$a_{n}=\frac{n-3}{3}$
(c)$a_{n}=3^{n}$
(d)$a_{n}=\frac{3n-2}{5}$
(e)$a_{n}=(-1^{n})2^{n}$
(f)$a_{n}=\fracn({n-2}){2}$
(g)$a_{n}=n^{2}-n+1$
(h)$a_{n}=2n^{2}-3n+1$
(i)$a_{n}=\frac{2n-3}{6}$

Answer 1

Answer with Step-by-step explanation:

(a) Given : an = 3n + 2

On putting n = 1

a1 = 3(1) + 2  

a1 = 3 + 2

a1 = 5

On putting n = 2

a2 = 3(2) + 2

a2 = 6 + 2

a2 = 8

On putting n = 3

a3 = 3(3) + 2  

a3 = 9 + 2

a3 = 11

On putting n = 4

a4 = 3(4) +2  

a4 = 12+ 2

a4 = 14

On putting n = 5

a5 = 3(5) + 2  

a5 = 15 + 2

a5 = 17

Hence, the first five terms of the following sequence are 5, 8, 11, 14, 17  

(b) Given : an = (n - 2)/3

On putting n = 1

a1= (1 - 2)/3

a1 = - ⅓  

On putting n = 2

a2 = (2 - 2)/3

a2 = 0

On putting n = 3

a3 = (3 - 2)/3

a3 = 1/3

On putting n = 4

a4 = (4 - 2)/3

a4  = 2/3

On putting n = 5

a5 = (5 - 2)/3

a5  = 3/3

a5 = 1

Hence, the first five terms of the following sequence are -⅓ , 0, 1/3, 2/3, 1

(c) Given : an = 3ⁿ

On putting n = 1

a1= 3¹  

a1 = 3

On putting n = 2

a2 = 3²  

a2 = 9

On putting n = 3

a3= 3³

a3 = 27

On putting n = 4

a4 = 3⁴

a4 = 81

On putting n = 5

a5 = 3⁵  

a5 = 243

Hence, the first five terms of the following sequence are 3, 9, 27, 81, 243

(d) an = (3n - 2)/5

On putting n = 1

a1= (3×1 - 2)/5

a1 = (3 -2)/5

a1 = 1/5  

On putting n = 2

a2 = (3×2 - 2)/5

a2 = (6 - 2)/5

a2 = 4/5  

On putting n  = 3

a3 = (3×3 - 2)/5

a3 = (9 - 2)/5

a3 = 7/5  

On putting n = 4

a4 = (3×4 - 2)/5

a4  = (12 -2)/5

a4 = 10/5  

a4 = 2

On putting n = 5

a5 = (3×5 - 2)/5

a5 = (15 -2)/5

a5 = 13/5  

Hence, the first five terms of the following sequence are ⅕, ⅘, 7/5, 2 , 13/5  

(e) Given : an = (-1)ⁿ . 2ⁿ

On putting n = 1

a1= (-1)¹. 2¹  

a1= (-1). 2  

a1 = - 2

On putting n = 2

a2 = (-1)². 2²  

a2 = (1). 4  

a2 = 4

On putting n  = 3

a3 = (-1)³ .2³

a3 = (-1). 8  

a3 = - 8

On putting n= 4

a4 = (-1)⁴.2⁴

a4 = (1).16  

a4 = 16

On putting n  = 5

a5= (-1)⁵ .2⁵

a5 = (-1).32  

a5 = - 32

Hence, the first five terms of the following sequence are -2, 4, -8, 16, -32

(f) Given : an = n(n - 2)/2

On putting n  = 1

a1=  1(1 - 2)/2

a1 = (1 × -1)/2

a1 = - ½  

On putting n = 2

a2 =  2(2 - 2)/2

a2 = 2(0)/2

a2 = 0

On putting n = 3

a3 =  3(3 - 2)/2

a3 = 3 (1)/2

a3 = 3/2  

On putting n  = 4

a4 =  4(4 - 2)/2

a4 = 4(2)/2 = 8/2

a4 = 4

On putting n = 5

a5 =  5(5 - 2)/2

a5  = 5(3)/2

a5 = 15/2

Hence, the first five terms of the following sequence are - ½ , 0, 3/2 ,4, 15/2  

(g) Given : an = n² - n + 1

On putting n= 1

a1= (1)² – 1 + 1  

a1 = 1 + 0

a1 = 1

On putting n = 2

a2 = (2)² - 2+1  

a2 = 4 - 1  

a2 = 3

On putting n= 3

a = (3)² - 3 + 1  

a3 = 9 - 2  

a3 = 7

On putting n= 4

a4 = (4)² - 4 + 1  

a4 = 16 - 3  

a4 = 13

On putting n = 5

a5 = (5)² – 5 + 1  

a5 = 25 - 4  

a5 = 21

Hence, the first five terms of the following sequence are 1, 3, 7, 13, 21.

(h) Given : an = 2n² - 3n + 1

On putting n = 1

a1= 2(1)²– 3(1) + 1  

a1 = 2 - 3 + 1  

a1 = 0

On putting n = 2

a2= 2(2)² – 3(2) + 1  

a2 = 8 - 6 + 1  

a2 = 3

On putting n = 3

a3 = 2(3)² – 3(3) + 1  

a3 = 18 - 9 + 1  

a3 = 10

On putting n = 4

a4 = 2(4)² – 3(4) + 1  

a4 = 32 -12 + 1  

a4 = 20 + 1

a4 = 21

On putting n= 5

a5 = 2(5)² – 3(5) + 1  

a5 = 50 - 15 + 1  

a5 = 36

Hence, the first five terms of the following sequence are 0,3,10,21,36.

(i) Given : an = (2n - 3)/6

On putting n = 1

a1=  (2 × 1 - 3)/6

a1= (2 - 3)/6

a1 = - ⅙  

On putting n = 2

a2 =  (2 × 2 - 3)/6

a2 = (4 - 3)/6

a2  =  ⅙  

On putting n = 3

a3 =  (2 × 3 - 3)/6

a3 = (6 - 3)/6

a3  = 3/6  

a3 = 1/2  

On putting n = 4

a4 =  (2 × 4 - 3)/6

a4 = (8 - 3)/6

a4  = 5/6

On putting n = 5

a5 =  (2 × 5 - 3)/6

a5 = (10 - 3)/6

a5  = 7/6  

Hence, the first five terms of the following sequence are -⅙,⅙, ½, ⅚, 7/6.