1. **State the problem:** We have an arithmetic sequence with terms $t(7) = 1056$ and $t(12) = 116$. We need to find $t(4)$. 2. **Recall the formula for the $n$-th term of an arithmetic sequence:** $$t(n) = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference. 3. **Write equations for the given terms:** $$t(7) = a + 6d = 1056$$ $$t(12) = a + 11d = 116$$ 4. **Subtract the first equation from the second to eliminate $a$:** $$a + 11d = 116$$ $$-(a + 6d = 1056)$$ \begin{aligned} & a + 11d - a - 6d = 116 - 1056 \\ & 5d = -940 \\ & d = \frac{-940}{5} = -188 \end{aligned} 5. **Substitute $d = -188$ back into one of the original equations to find $a$:** $$a + 6(-188) = 1056$$ $$a - 1128 = 1056$$ $$a = 1056 + 1128 = 2184$$ 6. **Find $t(4)$ using the formula:** $$t(4) = a + 3d = 2184 + 3(-188) = 2184 - 564 = 1620$$ **Final answer:** $$\boxed{1620}$$