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The sum of first $n$ terms of an A.P. is $2n^2 + 13n$. Find its $n^{th}$ term and hence $10^{th}$ term.
The sum of first $n$ terms of an A.P. is $2n^2 + 13n$. Find its $n^{th}$ term and hence $10^{th}$ term.
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Sol. $S_n = 2n^2 + 13n$
$S_1 = a_1 = 15$ (I Mark)
$S_2 = a_1 + a_2 = 34 \Rightarrow a_2 = 19$ (II Mark)
$\Rightarrow d = 19 - 15 = 4$ (III Mark)
$\therefore a_n = 15 + (n - 1) \times 4 = 4n + 11$ (IV Mark)
Hence $a_{10} = 4 \times 10 + 11 = 51$ (V Mark)
$S_1 = a_1 = 15$ (I Mark)
$S_2 = a_1 + a_2 = 34 \Rightarrow a_2 = 19$ (II Mark)
$\Rightarrow d = 19 - 15 = 4$ (III Mark)
$\therefore a_n = 15 + (n - 1) \times 4 = 4n + 11$ (IV Mark)
Hence $a_{10} = 4 \times 10 + 11 = 51$ (V Mark)