Determine which teqn of the arithmetic sequen a) \( 3 ; 5 ; 7 ; \ldots \) is equal to 27 b) \( -4 ; 1 ; 6 ; \ldots \) is equal to 56 c) \( 13 ; 10 ; 7 ; \ldots \) is equal to -44 d) \( 3 ; 3 \frac{1}{2} ; 4 ; \ldots \) is equal to \( 15 \frac{1}{2} \) e) \( -5 a ;-8 a ;-11 a ; \ldots \) is equal to \( -32 a \).

To determine the term of an arithmetic sequence that equals a certain value, we can use the formula for the n-th term of an arithmetic sequence: \[ a_n = a_1 + (n - 1)d \] where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number. a) In the sequence \(3 ; 5 ; 7 ; \ldots\): - First term \(a_1 = 3\) - Common difference \(d = 2\) - We want \(a_n = 27\): \[ 27 = 3 + (n - 1) \cdot 2 \] Rearranging gives \(n = 13\). b) In the sequence \(-4 ; 1 ; 6 ; \ldots\): - First term \(a_1 = -4\) - Common difference \(d = 5\) - We want \(a_n = 56\): \[ 56 = -4 + (n - 1) \cdot 5 \] Rearranging gives \(n = 12\). c) In the sequence \(13 ; 10 ; 7 ; \ldots\): - First term \(a_1 = 13\) - Common difference \(d = -3\) - We want \(a_n = -44\): \[ -44 = 13 + (n - 1) \cdot (-3) \] Rearranging gives \(n = 19\). d) In the sequence \(3 ; 3 \frac{1}{2} ; 4 ; \ldots\): - First term \(a_1 = 3\) - Common difference \(d = \frac{1}{2}\) - We want \(a_n = 15 \frac{1}{2}\): \[ 15.5 = 3 + (n - 1) \cdot \frac{1}{2} \] Rearranging gives \(n = 26\). e) In the sequence \(-5a ; -8a ; -11a ; \ldots\): - First term \(a_1 = -5a\) - Common difference \(d = -3a\) - We want \(a_n = -32a\): \[ -32a = -5a + (n - 1) \cdot (-3a) \] Rearranging gives \(n = 10\). Recapping: a) \(n = 13\) b) \(n = 12\) c) \(n = 19\) d) \(n = 26\) e) \(n = 10\)