Question

If sn, the sum of first n terms of an AP is given by sn= (3n2-4n), then find its nth term.

Answer 1

Answer:

$n^{th} \: term = (t_{n})=6n-7$

Step-by-step explanation:

$Given \\ Sum \: of \: first \: n \: terms \: of\\ \: an \: A.P \: (S_{n})= 3n^{2}-4n$

$Let \: the \: n^{th} \: term = (t_{n})$

$\boxed {t_{n}=S_{n}-S_{n-1}}$

$\implies t_{n}=3n^{2}-4n-[3(n-1)^{2}-4(n-1)]\\=3n^{2}-4n-[3(n^{2}-2n+1^{2}-4n+4]\\=3n^{2}-4n-3n^{2}+6n-3+4n-4\\=6n-7$

Therefore,

$n^{th} \: term = (t_{n})=6n-7$

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Answer 2

Answer:

Tn=6n-7

Step-by-step explanation:

Tn=3n²-4n-(3(n-1)²-4(n-1))

=3n²-4n-3(n²+1-2n)+4(n-1)

=3n²-4n-3n²-3+6n+4n-4

=6n-3-4

=6n-7

therefore, Tn =6n-7

hope it helps!!!

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