The sum of the first n terms of an AP is given by $S_{n} = 3 n^{2} - 4 n$ . Determine the AP and the $12^{t h}$ term.

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Solution

We have , $S_{n} = 3 n^{n}$ – 4n ---------------(i)

Replacing n by n-1 , we get

$S_{n - 1} = 3 (n - 1)^{2}$ – 4 (n-1) --------------(ii)

We know,

$a_{n} = S_{n} - S_{n - 1} = 3 n^{n} - 4 n - 3 (n - 1)^{2} - 4 (n - 1)$

$= 3 n^{2} - 4 n - 3 n^{2} + 3 - 6 n - 4 n + 4$

= $3 n^{2} - 4 n - 3 n^{2}$ – 3 +6n +4n -4 =6n -7

So, nth term $a_{n} = 6 n - 7$ ---------------(iii)

To get the AP, substitute n =1 , 2, 3,……. respectively in (iii) , we get

$a_{1}$ = 6 $\times$ 1 -7 = -1

$a_{2}$ =6 $\times$ 2-7 = 5

$a_{3}$ =6 $\times$ 3 – 7 =11

Also, to get $12^{t h}$ term, substitute n=12 in (iii), we get

$a_{12}$ = 6 $\times$ 12 – 7 = 65

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The sum of the first n terms of an AP is given by Sn=3n2−4n. Determine the AP and the 12th term.

The sum of the first n terms of an AP is given by $S_{n} = 3 n^{2} - 4 n$ . Determine the AP and the $12^{t h}$ term.