Question
Show that the sequence defined by $a_n = 3n^2 - 5$ is not an $A.P.$
✓
Answer
Given sequence is,
$a_n = 3n^2 - 5.$
$n^{th}$ term of given sequence $(a_n) = 3n^2 - 5.$
$(n + 1)^{th}$ term of given sequence $(a_n + 1) = 3(n + 1)^2 - 5$
$= 3(n^2 + 1^2 + 2n.1) - 5$
$= 3n^2 + 6n - 2$
$\therefore$ The common difference $(d) = a_n + 1 - an$
$d = (3n^2 + 6n - 2) - (3n^2 - 5)$
$= 3a^2 + 6n - 2 - 3n^2 + 5$
$= 6n + 3$
Common difference (d) depends on 'n' value
$\therefore$ Given sequence is not in A.P.
$a_n = 3n^2 - 5.$
$n^{th}$ term of given sequence $(a_n) = 3n^2 - 5.$
$(n + 1)^{th}$ term of given sequence $(a_n + 1) = 3(n + 1)^2 - 5$
$= 3(n^2 + 1^2 + 2n.1) - 5$
$= 3n^2 + 6n - 2$
$\therefore$ The common difference $(d) = a_n + 1 - an$
$d = (3n^2 + 6n - 2) - (3n^2 - 5)$
$= 3a^2 + 6n - 2 - 3n^2 + 5$
$= 6n + 3$
Common difference (d) depends on 'n' value
$\therefore$ Given sequence is not in A.P.
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