1. i) Three numbers are in G.P. Their product is 64 and sum is \( 124 / 5 \) find them. ii) If \( a, b, c \) are in GP and \( a^{x}=b^{y}=c^{2} \), prove that \( 1 / x+1 / z=2 / y \).

i) Let the three numbers be \( a, ar, ar^2 \), where \( r \) is the common ratio. Since they are in geometric progression, we have: \[ a \cdot ar \cdot ar^2 = 64 \] \[ a^3r^3 = 64 \] \[ (ar)^3 = 64 \] \[ ar = 4 \] The sum of the three numbers is: \[ a + ar + ar^2 = \frac{124}{5} \] \[ a(1 + r + r^2) = \frac{124}{5} \] We already know that \( ar = 4 \), so we can substitute \( a \) with \( \frac{4}{r} \): \[ \frac{4}{r}(1 + r + r^2) = \frac{124}{5} \] \[ 4(1 + r + r^2) = \frac{124}{5}r \] \[ 4 + 4r + 4r^2 = \frac{124}{5}r \] \[ 20 + 20r + 20r^2 = 124r \] \[ 20r^2 - 104r + 20 = 0 \] \[ 5r^2 - 26r + 5 = 0 \] This is a quadratic equation in \( r \). We can solve it using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ r = \frac{26 \pm \sqrt{(-26)^2 - 4 \cdot 5 \cdot 5}}{2 \cdot 5} \] \[ r = \frac{26 \pm \sqrt{676 - 100}}{10} \] \[ r = \frac{26 \pm \sqrt{576}}{10} \] \[ r = \frac{26 \pm 24}{10} \] We have two possible values for \( r \): \[ r = \frac{26 + 24}{10} = \frac{50}{10} = 5 \] \[ r = \frac{26 - 24}{10} = \frac{2}{10} = \frac{1}{5} \] If \( r = 5 \), then \( a = \frac{4}{r} = \frac{4}{5} \), and the three numbers are \( \frac{4}{5}, 4, 20 \). If \( r = \frac{1}{5} \), then \( a = \frac{4}{r} = 20 \), and the three numbers are \( 20, 4, \frac{4}{5} \). ii) Given that \( a, b, c \) are in geometric progression, we have: \[ b = ar \] \[ c = ar^2 \] Since \( a^x = b^y = c^2 \), we can write: \[ a^x = (ar)^y = (ar^2)^2 \] \[ a^x = a^{y+2} \] Since the bases are equal, the exponents must also be equal: \[ x = y + 2 \] Now, we want to prove that \( \frac{1}{x} + \frac{1}{z} = \frac{2}{y} \). We know that \( z = 2 \) because \( z = y + 2 \) and \( x = y + 2 \). So, we have: \[ \frac{1}{x} + \frac{1}{z} = \frac{1}{y + 2} + \frac{1}{2} \] To combine these fractions, we need a common denominator: \[ \frac{1}{y + 2} + \frac{1}{2} = \frac{2}{2(y + 2)} + \frac{y + 2}{2(y + 2)} \] \[ = \frac{2 + y + 2}{2(y + 2)} \] \[ = \frac{y + 4}{2(y + 2)} \] Since \( x = y + 2 \), we can substitute \( x \) for \( y + 2 \): \[ \frac{y + 4}{2(y + 2)} = \frac{y + 4}{2x} \] Now, we can divide both the numerator and the denominator by \( y \): \[ \frac{y + 4}{2x} = \frac{1 + \frac{4}{y}}{2} \] Since \( x = y + 2 \), we can substitute \( x \) for \( y +