The sum of the first n terms of an AP is given by $S n = (3 n^{2} - n)$ . Find its nth term.

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Solution

We know, $n^{t h}$ term $a_{n} = a + (n - 1) d$
Where
$a =$ First term
$d =$ Common difference
$n =$ number of terms
$a_{n} = n^{t h}$ term
Given,
$S_{n} = 3 n^{2} - n$
$S_{1} = 3 (1)^{2} - 1 = 3 - 1 = 2$
$S_{2} = 3 (2)^{2} - 2 = 12 - 2 = 10$
$a_{1} = 2$
$a_{2} = S_{2} - S_{1} = 10 - 2 = 8$

$a_{n} = a + (n - 1) d$
$= 2 + (n - 1) \times 6$
$= 2 + 6 n - 6$
$= 6 n - 4$ .

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The sum of the first n terms of an AP is given by Sn=(3n2−n). Find its nth term.

We know, nth term an=a+(n−1)dWherea= First termd= Common differencen= number of termsan=nth termGiven, Sn=3n2−nS1=3(1)2−1=3−1=2S2=3(2)2−2=12−2=10a1=2a2=S2−S1=10−2=8an=a+(n−1)d=2+(n−1)×6=2+6n−6=6n−4.

The sum of the first n terms of an AP is given by $S n = (3 n^{2} - n)$ . Find its nth term.