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rev | Admin | (Created page with "'''Solution: D''' Let <math>P=</math> price of bond <math>1(=</math> price of bond 2<math>) .8</math> years means 16 periods. Bond 1 gives <math>P=F v^{16}+F r a_{\overline{16} \mid}=F(1.025)^{-16}+F(.03) a_{\overline{16} \mid .025}</math>. For bond 2, let <math>n</math> be the number of years. There are <math>2 n</math> periods. Bond 2 gives <math>P=F(1.025)^{-2 n}+F(.0275) a_{\overline{2 n \mid} .025}</math>. Equating the two expressions and cancelling <math>F</math>...") | Nov 29'23 at 15:50 | +1,303 |