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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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