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Question
If Sn.., the sum of first n terms of an AP is given by Sn =3n2- 4n, find the nth term.
Solution
sn-=3n2 - 4n
Let Sn-1 be sum of (n - 1) terms
tn = Sn - Sn-1
(3n2 - 4n) - [3(n - 1)2- 4(n- 1)]
=(3n2- 4n) - [3n2 - 6n + 3 - 4n+4]
= 3n2 - 4n - 3n2 + 10n- 7
: . tn =6n - 7
Therefore, required nth term =60- 7
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