if sn =3n2 - 4, find sum of first 10 terms
Answers
Answered by
1
Answer:
Sn=3n2+4
n=10
S10=3×10×10+4
S10=304
Answered by
0
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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