(m) The sum of the first 5 terms of a convergent geometric series is 62 and the sum to infinity of the series is 64. Determine the sum of the first 10 terms.

Let the first term of the geometric series be \( a \) and the common ratio be \( r \). The formula for the sum of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \] For the first 5 terms: \[ S_5 = a \frac{1 - r^5}{1 - r} = 62 \] The sum to infinity \( S_\infty \) is given by: \[ S_\infty = \frac{a}{1 - r} = 64 \] From the second equation, we can express \( a \) in terms of \( r \): \[ a = 64(1 - r) \] Now, substituting \( a \) back into the first equation: \[ 62 = 64(1 - r) \frac{1 - r^5}{1 - r} \] This simplifies to: \[ 62 = 64(1 - r^5) \] Now, solving for \( r^5 \): \[ 62 = 64 - 64r^5 \] \[ 64r^5 = 64 - 62 \] \[ 64r^5 = 2 \] \[ r^5 = \frac{2}{64} = \frac{1}{32} \] Taking the fifth root: \[ r = \left(\frac{1}{32}\right)^{1/5} \] \[ r = \frac{1}{2} \] Now, substitute \( r \) back into the equation for \( a \): \[ a = 64(1 - \frac{1}{2}) = 64 \cdot \frac{1}{2} = 32 \] Thus, \( a = 32 \) and \( r = \frac{1}{2} \). Now we can calculate the sum of the first 10 terms \( S_{10} \): \[ S_{10} = a \frac{1 - r^{10}}{1 - r} = 32 \frac{1 - \left(\frac{1}{2}\right)^{10}}{1 - \frac{1}{2}} \] Calculating \( \left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \): \[ S_{10} = 32 \frac{1 - \frac{1}{1024}}{\frac{1}{2}} = 32 \cdot 2 \left(1 - \frac{1}{1024}\right) \] \[ = 64 \left(1 - \frac{1}{1024}\right) = 64 \cdot \frac{1023}{1024} \] \[ = \frac{64 \cdot 1023}{1024} = \frac{65536}{1024} - \frac{64}{1024} = \frac{65536 - 64}{1024} = \frac{65536 - 64}{1024} = \frac{65536 - 64}{1024} = 63.9375 \] Thus, the sum of the first 10 terms is: \[ \boxed{63.9375} \]