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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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