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If the numbers $\mathrm{n}-3,4 \mathrm{n}-2,5 \mathrm{n}+1$ are in AP, what is the value of $\mathrm{n} ?$
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$\begin{aligned} & \because(\mathrm{n}-3), 4 \mathrm{n}-2,5 \mathrm{n}+1 \text { are in A.P. } \\ & \Rightarrow(4 \mathrm{n}-2)-(\mathrm{n}-3)=(5 \mathrm{n}+1)-(4 n-2) \\ & \Rightarrow 3 \mathrm{n}+1=\mathrm{n}+3 \\ & \Rightarrow \mathrm{n}=1 \end{aligned}$
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