Question

Which of the following can be the nth term of an A.P.? (1) 4/n+2 (2) 4n + 5 (3) 2n^2 + 3 (4) n^2– 5n

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{(2) \ 4n+5 \ is \ the \ correct \ option.}$

$\sf{4n+5 \ can \ be \ the \ n^{th} \ term \ of}$

$\sf{an \ AP.}$

$\sf\pink{To \ find;}$

$\sf{The \ n^{th} \ term \ of \ the \ AP.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{(1) \ In \ \frac{4}{n}+2}$

$\sf{Let \ n=1,}$

$\sf{\therefore{t_{1}=4+2=6}}$

$\sf{Let \ n=2,}$

$\sf{\therefore{t_{2}=\frac{4}{2}+2=2+2=4}}$

$\sf{Let \ n=3,}$

$\sf{\therefore{t_{3}=\frac{4}{3}+2=\frac{10}{3}}}$

$\sf{Here,}$

$\sf{t_{2}-t_{1}=6-4=2...(1)}$

$\sf{t_{3}-t_{2}=\frac{10}{3}-6=\frac{-8}{3}...(2)}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{t_{2}-t_{1} \ \neq \ t_{3}-t_{2}}$

$\sf{\therefore{It \ will \ not \ form \ an \ arithmetic}}$

$\sf{progression.}$

_______________________________

$\sf{(2) \ In \ 4n+5}$

$\sf{Let \ n=1,}$

$\sf{\therefore{t_{1}=4+5=9}}$

$\sf{Let \ n=2,}$

$\sf{\therefore{t_{2}=4(2)+5=13}}$

$\sf{Let \ n=3,}$

$\sf{\therefore{t_{3}=4(3)+5=17}}$

$\sf{t_{2}-t_{1}=13-9=4...(1)}$

$\sf{t_{3}-t_{2}=17-13=4...(2)}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{t_{2}-t_{1}=t_{3}-t_{2}}$

$\sf{Hence, \ the \ common \ difference}$

$\sf{is \ constant \ and \ arithmetic \ progression}$

$\sf{can \ be \ formed.}$

_______________________________

$\sf{(3) \ In \ 2n^{2}+3}$

$\sf{Let \ n=1,}$

$\sf{\therefore{t_{1}=2+3=5}}$

$\sf{Let \ n=2,}$

$\sf{\therefore{t_{2}=2\times(2)^{2}+3=11}}$

$\sf{Let \ n=3,}$

$\sf{\therefore{t_{3}=2\times(3)^{2}+3=21}}$

$\sf{t_{2}-t_{1}=11-5=6...(1)}$

$\sf{t_{3}-t_{2}=21-11=10...(2)}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{t_{2}-t_{1} \ \neq \ t_{3}-t_{2}}$

$\sf{\therefore{It \ will \ not \ form \ an \ arithmetic}}$

$\sf{progression.}$

_______________________________

$\sf{(4) \ In \ n^{2}-5n}$

$\sf{Let \ n=1,}$

$\sf{\therefore{t_{1}=1-5=-4}}$

$\sf{Let \ n=2,}$

$\sf{\therefore{t_{2}=(2)^{2}-5(2)=4-10=-6}}$

$\sf{Let \ n=3,}$

$\sf{\therefore{t_{3}=(3)^{2}-5(3)=9-27=-16}}$

$\sf{t_{2}-t_{1}=-6-(-4)=-2...(1)}$

$\sf{t_{3}-t_{2}=-16-(-6)=10...(2)}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{t_{2}-t_{1} \ \neq \ t_{3}-t_{2}}$

$\sf{\therefore{It \ will \ not \ form \ an \ arithmetic}}$

$\sf{progression.}$

Answer 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

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