ARITHMETIC GEOMETRIC AND HARMONIC PROGRESSION

Subscribe to our ā–¶ļø YouTube channel šŸ”“ for the latest videos, updates, and tips.

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

t= 5/7

n  =  2

  t= 7/10

n  =  3

  t= 9/13

n  =  4

  t= 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

t= 1/1 = 1

n  =  2

  t= 4/3

n  =  3

  t= 7/9

n  =  4

  t= 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.

Subscribe to our ā–¶ļø YouTube channel šŸ”“ for the latest videos, updates, and tips.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

About Us  |  Contact Us  |  Privacy Policy

©All rights reserved. onlinemath4all.com

onlinemath4all_official_badge1.png

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 40)

    Mar 03, 26 06:53 PM

    digitalsatmath36.png
    Digital SAT Math Problems and Solutions (Part - 40)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 39)

    Mar 03, 26 04:59 PM

    digitalsatmath243.png
    Digital SAT Math Problems and Solutions (Part - 39)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 38)

    Mar 03, 26 10:05 AM

    digitalsatmath252.png
    Digital SAT Math Problems and Solutions (Part - 38)

    Read More