Find the number of terms in the arithmetic sequence \( -5 ;-11 ;-17 ; \ldots ;-491 \) How many terms are there in the arithmetic sequence \( -2 ; \frac{1}{2} ; 3 ; \ldots ; \frac{361}{2} \) ?

To find the number of terms in the first arithmetic sequence \( -5, -11, -17, \ldots, -491 \), we first identify the first term \( a = -5 \) and the common difference \( d = -11 - (-5) = -6 \). The \( n \)-th term of an arithmetic sequence can be expressed as: \[ a_n = a + (n-1)d \] Setting \( a_n = -491 \): \[ -491 = -5 + (n-1)(-6) \] Now, let's solve for \( n \): \[ -491 + 5 = (n-1)(-6) \] \[ -486 = (n-1)(-6) \] Dividing both sides by -6: \[ n-1 = \frac{-486}{-6} = 81 \] Thus, \[ n = 81 + 1 = 82 \] So there are **82 terms** in the first arithmetic sequence. For the second arithmetic sequence \( -2, \frac{1}{2}, 3, \ldots, \frac{361}{2} \), we identify the first term \( a = -2 \) and the common difference \( d = \frac{1}{2} - (-2) = \frac{1}{2} + 2 = \frac{5}{2} \). Setting \( a_n = \frac{361}{2} \): \[ \frac{361}{2} = -2 + (n-1)\left(\frac{5}{2}\right) \] Converting -2 into an equivalent fraction: \[ -2 = \frac{-4}{2} \] Now, substituting this in: \[ \frac{361}{2} = \frac{-4}{2} + (n-1)\left(\frac{5}{2}\right) \] \[ \frac{361}{2} + \frac{4}{2} = (n-1)\left(\frac{5}{2}\right) \] \[ \frac{365}{2} = (n-1)\left(\frac{5}{2}\right) \] Multiplying both sides by 2 to eliminate the fraction: \[ 365 = (n-1) \cdot 5 \] Dividing both sides by 5: \[ n-1 = \frac{365}{5} = 73 \] Thus, \[ n = 73 + 1 = 74 \] So there are **74 terms** in the second arithmetic sequence.