ARITHMETIC GEOMETRIC AND HARMONIC PROGRESSION

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Question :

Write the first 6 terms of the sequences whose nth terms are given below and classify them as arithmetic progression, geometric progression, arithmetico-geometric progression, harmonic progression and none of them.

(i)  1/2n+1

(ii)  (n + 1)(n + 2) / (n + 3)(n + 4)

(iii) 4 (1/2)n

(iv) (āˆ’1)n/n

(v) (2n+3) / (3n+4)

(vi) 2018

(vii) (3nāˆ’2)/(3nāˆ’1)

Solution :

(i)  1/2n+1

Let tn  =  1/2n+1

n  =  1

t1  =  1/21+1 

=  1/4  

n  =  2

t2  =  1/22+1 

=  1/8  

n  =  3

t3  =  1/23+1 

=  1/16  

n  =  4

t4  =  1/24+1 

=  1/32  

n  =  5

t5  =  1/25+1 

=  1/64  

n  =  6

t6  =  1/26+1 

=  1/128  

First 6 terms of the sequence are,

1/4, 1/8, 1/16, 1/32, 1/64, 1/128,............

Since the common ratio is same, it is GP.

(ii)  (n + 1)(n + 2) / (n + 3)(n + 4)

Solution :

Let tn  =  (n + 1)(n + 2) / (n + 3)(n + 4)

n  =  1

t1 = (1+1)(1+2)/(1+3)(1+4)

=  6/20

n  =  2

t1 = (2+1)(2+2)/(2+3)(2+4)

=  12/30

n  =  3

t3 = (3+1)(3+2)/(3+3)(3+4)

=  20/42

n  =  4

t4 = (4+1)(4+2)/(4+3)(4+4)

=  30/56

n  =  5

t3 = (5+1)(5+2)/(5+3)(5+4)

=  42/72

n  =  6

t3 = (6+1)(6+2)/(6+3)(6+4)

=  56/90

First 6 terms of the sequence are,

6/20, 12/30, 20/42, 30/56, 42/72, 56/90,............

Common difference is not same, so it is not A.P

Common ratio is not same, so it is not G.P

It is not H.P

Hence the answer is none of them.

(iii) 4 (1/2)n

Solution :

Let tn  =  4 (1/2)n

n  =  1

t1  =  4 (1/2)n  

=  4(1/2)1

=  2

n  =  2

t2  =  4 (1/2)n  

=  4(1/2)2

=  1

n  =  3

t3  =  4 (1/2)n  

=  4(1/2)3

=  1/2

n  =  4

t4  =  4 (1/2)4  

=  1/4

n  =  5

t5  =  4 (1/2)5  

=  1/8

n  =  6

t6  =  4 (1/2)6  

=  1/16

First 6 terms of the sequence are,

2, 1, 1/2, 1/4, 1/8, 1/16,....................

The common ratio is same, so it is G.P

(iv) (āˆ’1)n/n

Solution :

n  =  1

t1  =  -1/1

=  -1

n  =  2

t2  =  1/2

n  = 3

t2  =  -1/3

n  = 4

t4  =  1/4

n  =  5

t5  =  -1/5

n  =  6

t6  =  1/6

First 6 terms of the sequence are,

-1, 1/2, -1/3, 1/4, -1/5, 1/6, .................

Common difference is not same, so it is not A.P

Common ratio is not same, so it is not G.P

It is not H.P

Hence the answer is none of them.

(v) (2n+3) / (3n+4)

Solution :

Let tn  =  (2n+3) / (3n+4)

n  =  1

t1 = 5/7

n  =  2

  t2 = 7/10

n  =  3

  t3 = 9/13

n  =  4

  t4 = 11/16

n  =  5

t5  =  13/19

n  =  6

t6  =  15/22

First 6 terms of the sequence are,

5/7, 7/10, 9/13, 11/16, 13/19, 15/22,....................

Hence the answer is none of these.

(vi) 2018

Solution :

The answer is none of these.

(vii) (3nāˆ’2)/(3nāˆ’1)

Solution :

Let tn  =  (3nāˆ’2)/(3nāˆ’1)

n  =  1

t1 = 1/1 = 1

n  =  2

  t2 = 4/3

n  =  3

  t3 = 7/9

n  =  4

  t4 = 10/27

n  =  5

t5  =  13/81

n  =  6

t6  =  16/243

First 6 terms of the sequence are,

1, 4/3, 7/9, 10/27, 13/81, 16/243, .............

Hence the answer is arithmetico-geometric progression.

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