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# Justify whether it is true to say that the following are the $ n^{\text {th }} $ terms of an AP.

$ 1+n+n^{2} $

Given:

$a_n = 1 + n + n^2$

To do:

We have to justify whether it is true to say that $a_n = 1 + n + n^2$ is the \( n^{\text {th }} \) term of an AP.

Solution:

To check whether the sequence defined by $a_n = 1 + n + n^2$ is an A.P., we have to check whether the difference between any two consecutive terms is equal.

Let us find the first few terms of the sequence by substituting $n=1, 2, 3....$

When $n=1$,

$a_1=1+1+(1)^2$

$=1+1+1$

$=3$

$a_2=1+2+(2)^2$

$=3+4$

$=7$

$a_3=1+3+(3)^2$

$=4+9$

$=13$

$a_4=1+4+(4)^2$

$=5+16$

$=21$

Here,

$a_2-a_1=7-3=4$

$a_3-a_2=13-7=6$

$d=a_4-a_3=21-13=8$

$a_2-a_1≠a_3-a_2≠a_4-a_3$

Hence, the given sequence is not an A.P.

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