Show that the list of numbers defined by $a_{n} = 3 n^{2} - 5$ is not an AP.

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Solution

A sequence is said to be in A.P if the difference between $2$ connective terms is the same

$a_{n} = 3 n^{2} - 5$

$a_{n + 1} = 3 (n + 1)^{2} - 5 = 3 n^{2} + 6 n + 3 - 5$
$= 3 n^{2} + 6 n - 2$

$a_{n + 1} - a_{n} = (3 n^{2} - 5) - (3 n^{2} + 6 n - 2)$
$= - 5 - 6 n + 2$
$= - 6 n - 3$

Difference between terms changes with "n", it is not constant, therefore given sequence is not in A.P.

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