Math, asked by misseemasingh1975, 9 months ago

which of the following can be nth term of an ap 4/n+2, 4n+5, 2n^2+3, n^2-5n with reason briefly​

Answers

Answered by DevendraLal
2

Given:

Series with a notation of the n term 4/n+2, 4n+5, 2n^2+3, n^2-5n

To find:

Which of the given term will be the nth term of the AP.

Solution:

1) By putting the value of n as 1, 2, 3 and so on we get the series:

Series for 4/n+2:

  • 4/3, 4/4, 4/5, 4/6, ............

the common difference for the terms:

  • -1/3, -1/5 (which is not same)

Series for 4n+5

  • 9, 13, 17, 21,...........

The common difference for the terms :

  • 4, 4, 4, 4............. ( same for all terms)

Series for 2n^2+3

  • 5, 11, 21, 35..........

The common difference for the terms :

  • 6, 10, 14, ........... ( different)

Series for n^2-5n

  • -4, -6, -6,.........

The common difference for the terms :

  • -2, 0,.....(different)

Series for 4n+5 can be the nth term of an AP.

Answered by amitnrw
2

Given :  (1) 4/n+2 (2) 4n + 5 (3) 2n^2 + 3 (4) n^2– 5n

To find : Which of the given terms can be the nth term of an A.P

Solution :

Common difference  of an AP =  aₙ₊₁ - aₙ

Common difference  should not depend upon n  for an AP as its fixed and independent of n

Lets check each

4/n+2

d =  4/(n + 3) - 4/(n + 2)

=> d = (4n + 8  - 4n - 12)/(n+3)(n+2)

=> d =  -4/(n+3)(n+2)

dependent upon n

Hence can not be an AP

4n + 5

d = 4(n+1) + 5 - (4n + 5)

=> d = 4

hence this can be an AP

hence 4n + 5 can be nth term of an AP

2n² + 3

d = 2(n+1)² + 3 -  (2n² + 3)

=> d = 2n²  + 4n + 2 + 3  - 2n²   - 3

=> d = 4n + 2

Dependent upon n

can not be an AP

n² - 5n

=> d = (n+1)² -5(n+1)  - (n² - 5n)

=> d = n² + 2n + 1 - 5n - 5     -n² + 5n

=> d = 2n - 4

Dependent upon n

can not be an AP

4n + 5 can be the nth term of an AP

Learn more:

the seventeenth term of an ap is 5 morethan twice its eighteenth ...

https://brainly.in/question/8045489

if mth term of an A.P.is n and nth term is m, show that (m+n)th term ...

https://brainly.in/question/1085792

Similar questions