Show that the sequence defined by $a_n = 3n^2 – 5$ is not an A.P.



Given:

$a_n = 3n^2 – 5$ 

To do:

We have to show that the sequence defined by $a_n = 3n^2 – 5$ is not an A.P.

Solution:

To  show that the sequence defined by $a_n = 3n^2 – 5$ is not an A.P., we have to show that the difference between any two consecutive terms is not equal.

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

When $n=1$,

$a_1=3(1)^2-5$

$=3(1)-5$

$=3-5$

$=-2$

$a_2=3(2)^2-5$

$=3(4)-5$

$=12-5$

$=-7$

$a_3=3(3)^2-5$

$=3(9)-5$

$=27-5$

$=22$

$a_4=3(4)^2-5$

$=3(16)-5$

$=48-5$

$=43$

Here,

$a_2-a_1=-7-(-2)=-7+2=-5$

$a_3-a_2=22-(-7)=22+7=29$

$a_4-a_3=43-22=21$

$a_2-a_1≠a_3-a_2≠a_4-a_3$

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

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