12. If Sn the sum of n terms of an A.P. is given by Sn = 3n2 – 4n, find the nth term.
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Sn = 3n^2-4n
S(n-1) = 3(n-1)^2-4(n-1)
Sn- S(n-1)= (3n^2-4n) - {3(n-1)^2-4(n-1)}
an= (3n^2-4n) -{3(n^2-2n+1)- 4n+4}
= 3n^2-4n-(3n^2-6n+3-4n+4)
= 3n^2-4n-(3n^2-10n+7)
=3n^2-4n-3n^2+10n-7
=6n-7
nth term is 6n-7
S(n-1) = 3(n-1)^2-4(n-1)
Sn- S(n-1)= (3n^2-4n) - {3(n-1)^2-4(n-1)}
an= (3n^2-4n) -{3(n^2-2n+1)- 4n+4}
= 3n^2-4n-(3n^2-6n+3-4n+4)
= 3n^2-4n-(3n^2-10n+7)
=3n^2-4n-3n^2+10n-7
=6n-7
nth term is 6n-7
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