Justify whether it is true to say that the following are the nth terms of an AP.
(i) `2n-3 " " ` (ii) `3n^(2)+5 " " ` (iii) `1+n+n^(2)`

Explanation

(i) Yes, here `a_(n)=2n-3`
Put`n=1, " " a_(1)=2(1)-3= -1`
Put`n=2, " " a_(2)=2(2)-3= 1`
Put`n=3, " " a_(3)=2(3)-3= 3`
Put`n=4, " " a_(4)=2(4)-3= 5`
List of numbers becomes `-1,1,3, …`
Here, `a_(2)-a_(1)=1-(-1)=1+1=2`
`a_(3)-a_(2)=3-1=2`
`a_(4)-a_(3)=5-3=2`
` :' a_(2)-a_(1)=a_(3)-a_(2)=a_(4)-a_(3)=...`
Hence, `2n-3` is the nth term of an AP.
(ii) No, here `a_(n)=3n^(2)+5`
Put`n=1, " " a_(1)=3(1)^(2)+5=8`
Put`n=2, " " a_(2)=3(2)^(2)+5=3(4)+5=17`
Put`n=3, " " a_(3)=3(3)^(2)+5=3(9)+5= 32`
So, the list of number becomes `8,17,32, …`
Here, `a_(2)-a_(1)=17-8=9`
`a_(3)-a_(2)=32-17=15`
` :. a_(2)-a_(1) ne a_(3)-a_(2)`
Since, the successive difference of the list is not same. So, it does not form an AP.
(iii) No, here `a_(n)=1+n+n^(2)`
Put`n=1, " " a_(1)=1+1+(1)^(2)=3`
Put`n=2, " " a_(2)=1+2+(2)^(2)=1+2+4=7`
Put`n=3, " " a_(3)=1+3+(3)^(2)=1+3+9= 13`
So, the list of number becomes `3,7,13, …`
Here, `a_(2)-a_(1)=7-3=4`
`a_(3)-a_(2)=13-7=6`
` :. a_(2)-a_(1) ne a_(3)-a_(2)`
Since, the successive difference of the list is not same. So, it does not form an AP.