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If the sum of ' $n$ ' terms of an arithmetic progression is $n^{2}-2 n$, then what is the $n^{\text {th }}$ term?
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The correct answer is:
$2 n-3$
Given,
$\begin{aligned} & \mathrm{S}_{\mathrm{n}}=\mathrm{n}^{2}-2 \mathrm{n} \\ \therefore & \quad \mathrm{ a}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1} \\=& \mathrm{n}^{2}-2 \mathrm{n}-\left[(\mathrm{n}-1)^{2}-2(\mathrm{n}-1)\right] \\=& \mathrm{n}^{2}-2 \mathrm{n}-\left[\mathrm{n}^{2}+1-2 \mathrm{n}-2 \mathrm{n}+2\right]=2 \mathrm{n}-3 \end{aligned}$
$\begin{aligned} & \mathrm{S}_{\mathrm{n}}=\mathrm{n}^{2}-2 \mathrm{n} \\ \therefore & \quad \mathrm{ a}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1} \\=& \mathrm{n}^{2}-2 \mathrm{n}-\left[(\mathrm{n}-1)^{2}-2(\mathrm{n}-1)\right] \\=& \mathrm{n}^{2}-2 \mathrm{n}-\left[\mathrm{n}^{2}+1-2 \mathrm{n}-2 \mathrm{n}+2\right]=2 \mathrm{n}-3 \end{aligned}$
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