\( \left. \begin{array} { l l } { a = ? } & { T 7 = a + ( 7 - 1 ) d } \\ { d = 7 } & { T 7 = a + 6 d = - 20 } \\ { d = ? } & { } \\ { } & { 120 = a t } \end{array} \right. \)

We start with the information given for an arithmetic sequence. The formulas and data are as follows: 1. The nth‐term formula for an arithmetic sequence is   Tn = a + (n – 1)d. In particular, for the 7th term we have:   T7 = a + (7 – 1)d = a + 6d. 2. You are told that d = 7 and that T7 = –20. In other words,   a + 6(7) = –20. Step 1. Solve for a:   a + 42 = –20   => a = –20 – 42   => a = –62. So the first term a is –62, and the common difference d is 7. Step 2. Interpreting 120 = at There is one more equation given:   120 = a t. Since we now know a = –62, we can solve for t (interpreting “t” as some unknown quantity):   120 = (–62)t   => t = 120 / (–62)   => t = –60/31  (after dividing numerator and denominator by 2). Thus, the unknown t is –60/31. Final Answers:   First term, a = –62,   Common difference, d = 7,   and from 120 = a·t, we get t = –60/31. If you interpret “120 = at” in another context, you would need to adjust the answer accordingly, but with the given information, these are the solutions.