Determine \( m \) in each of the geometric series: a) \( 3+6+12+\ldots \) (to \( m \) terms) \( =93 \)

To find \( m \) in the geometric series \( 3 + 6 + 12 + \ldots \) (to \( m \) terms) that equals 93, we start by identifying the first term \( a = 3 \) and the common ratio \( r = \frac{6}{3} = 2 \). The formula for the sum \( S_n \) of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{r^n - 1}{r - 1} \] Substituting the known values: \[ S_m = 3 \frac{2^m - 1}{2 - 1} = 3(2^m - 1) \] Setting this equal to 93: \[ 3(2^m - 1) = 93 \] Dividing both sides by 3: \[ 2^m - 1 = 31 \] Adding 1 to both sides: \[ 2^m = 32 \] Recognizing that \( 32 = 2^5 \): \[ m = 5 \] Thus, the value of \( m \) is \( 5 \).