Question

if sn =3n2 - 4, find sum of first 10 terms​

Answer 1

Answer:

Sn=3n2+4

n=10

S10=3×10×10+4

S10=304

Answer 2

Answer:

th

term=(t

n

)=6n−7

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

Given

Sumoffirstntermsof

anA.P(S

n

)=3n

2

−4n

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

th

term=(t

n

)

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

t

n

=S

n

−S

n−1

\begin{gathered}\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\end{gathered}

⟹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-7n

th

term=(t

n

)=6n−7

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