If \(\frac{37}{13} = 2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}},\) where x

1.	If \(\frac{37}{13} = 2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}},\) where x, y, z are natural numbers, then x, y, z are(a) 1, 2, 5 (b) 1, 5, 2 (c) 5, 2, 11 (d) 11, 2, 5
Answer» (b) 1, 5, 2 \(= 2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{37}{13} = 2\frac{11}{13}= 2+\frac{11}{13}\) \(⇒\frac{1}{x+\frac{1}{y+\frac{1}{z}}}= \frac{11}{13}⇒x+\frac{1}{y+\frac1z}=\frac{13}{11}\) ⇒ \(x+\frac{1}{y+\frac1z}=1+\frac2{11}\) ⇒ x = 1, y = \(\frac1z=\frac{11}{2}=5\frac12=5+\frac12\) ⇒ x = 1, y = 5, z = 2

If \(\frac{37}{13} = 2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}},\) where x, y, z are natural numbers, then x, y, z are(a) 1, 2, 5 (b) 1, 5, 2 (c) 5, 2, 11 (d) 11, 2, 5

Answer»

(b) 1, 5, 2

\(= 2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{37}{13} = 2\frac{11}{13}= 2+\frac{11}{13}\)

\(⇒\frac{1}{x+\frac{1}{y+\frac{1}{z}}}= \frac{11}{13}⇒x+\frac{1}{y+\frac1z}=\frac{13}{11}\)

⇒ \(x+\frac{1}{y+\frac1z}=1+\frac2{11}\)

⇒ x = 1, y = \(\frac1z=\frac{11}{2}=5\frac12=5+\frac12\)

⇒ x = 1, y = 5, z = 2