If for a sequence $(t_{n}), S_{n} = 4 n^{2} - 3 n$ , show that the sequence is an $A . P .$

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Solution

$S_{n} = 4 n^{2} - 3 n$
Now, $S_{n + 1} = 4 (n + 1)^{2} - 3 (n + 1)$
We know that, $t_{n + 1} = S_{s + 1} - S_{n}$
$t_{n + 1} = 4 n^{2} + 5 n + 1 - 4 n^{2} + 3 n$
$t_{n + 1} = 8 n + 1$
Thus $t_{n + 1}$ is linear equation of $n$
hence it is an $A . P$
Now $t_{n + 1} = 8 n + 1$
comlplaning with $a_{n + 1} = a + (n - 1)$
$a = 1$ and $d = 8$

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