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

i)

Solution.

$\\Here a\textsubscript{n} = 2n - 3\\ Put n = 1, a\textsubscript{1} =2(1) - 3 = - 1\\ Putn=2, a\textsubscript{2} =2(2) - 3 = 4 - 3 = 1\\ Putn=3, a\textsubscript{3}=2(3) - 3 = 6 - 3 = 3\\ Putn=4, a\textsubscript{4} =2(4) - 3 = 8 - 3 = 5\\ Numbers are : -1, 1, 3, 5, \ldots ..\\ a\textsubscript{2}- a\textsubscript{1} = 1 - (-1) = 1 + 1 = 2\\ a\textsubscript{3}- a\textsubscript{2} = 3 - 1 = 2\\ a\textsubscript{4}- a\textsubscript{3} = 5 - 3 = 2\\$

Hence, 2n – 3 is the nth term of an AP because the common difference is the same.

ii)

Solution.

$\\Here, a\textsubscript{n} = 3n\textsuperscript{2} + 5\\ Put n = 1, a\textsubscript{1} =3(1)\textsuperscript{2} + 5 = 3 + 5 = 8\\ Put n =2, a\textsubscript{2} =3(2)\textsuperscript{2} + 5 = 12 + 5 = 17\\ Put n = 3, a\textsubscript{3} =3(3)\textsuperscript{2} + 5 = 27 + 5 = 32\\ Numbers are: 8, 17, 32 \ldots \\ a\textsubscript{2} - a\textsubscript{1} = 17 - 8 = 9\\ a\textsubscript{3}- a\textsubscript{2} = 32 - 17 = 15\\$

Here the common difference is not the same

Therefore series $3n^2+5$ is not the nth term of an AP.

iii)

Solution.

$\\Here a\textsubscript{n} = 1 + n + n\textsuperscript{2}\\ Put n = 1, a\textsubscript{1} = 1 + 1 + (1)\textsuperscript{2} = 3\\ Put n = 2, a\textsubscript{2} = 1 + 2 + (2)\textsuperscript{2} = 7\\ Put n = 3, a\textsubscript{3} = 1 + 3 + (3)\textsuperscript{2} = 13\\ The numbers are 3, 7, 13 \ldots \\ a\textsubscript{2} - a\textsubscript{1} = 7 - 3 = 4\\ a\textsubscript{3}- a\textsubscript{2} = 13 - 7 = 6\\$

Here the common difference is not the same

$\text{Hence}$ $1 + n + n^2$  $\text{is not the nth term of an AP.}$