The sum of the first $n$ terms of an AP is given by $S_{n} = (3 n^{2} - 4 n) .$ Find its (i) $n^{t h}$ term, (ii) first term, and (iii) common difference.

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Solution

Given:
$S_{n} = 3 n^{2} - 4 n$
Now, Sum of one term $= S_{1} = 3 \times 1^{2} - 4 \times 1 = 3 - 4 = - 1$
$\Rightarrow S_{1} = a_{1} = a = - 1 \dots \dots (i)$
Hence, the first term is $- 1.$
Sum of two terms $= S 2 = 3 \times 2^{2} - 4 \times 2 = 12 - 8 = 4$
$\Rightarrow S_{2} = a_{1} + a_{2} = 4 \dots \dots (i i)$
Subtracting eq.(i) from (ii), we get
$a_{2} = 5$
$\Rightarrow a_{2} = a + d = 5$ [ $∵ a = - 1$ ]
$\Rightarrow d = 6$
Hence, the common difference is $6$ .
Therefore, $n^{t h}$ term is given by $a_{n} = a + (n - 1) d$
$a_{n} = - 1 + (n - 1) 6 = - 1 + 6 n - 6 = 6 n - 7$
Hence, the $n^{t h}$ term is $6 n - 7$
Hence,
(i) $n^{t h}$ term is $(6 n - 7)$
(ii) first term is $- 1$
(iii) common difference is $6$

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