Question

Sum of first 4 term $S_{4} = 6$

$$\begin{array}{r} \frac{4}{2}(2a + 3d) = 6 \Rightarrow 2a + 3d = 3\#(1) \end{array}$$

Sum of first 6 terms $S_{6} = 4$

$$\begin{array}{r} \frac{6}{2}(2a + 5d) = 4 \Rightarrow 2a + 5d = \frac{4}{3}\ldots\#(2) \\ \text{~}\text{eq.}\text{~}(2) - \text{~}\text{eq.}\text{~}(1)\#(2) \\ \ (2a + 5d) - (2a + 3d) = \frac{4}{3} - 3\#(2) \\ \ \Rightarrow d = - \frac{5}{6}\#(2) \\ \ \therefore 2a + 3\left( - \frac{5}{6} \right) = 3 \Rightarrow a = \frac{11}{4}\#(2) \\ S_{12} = \frac{12}{2}\left\{ 2 \times \frac{11}{4} + (12 - 1)\left( - \frac{5}{6} \right) \right\}\#(2) \\ S_{12} = 6\left( - \frac{22}{6} \right) = - 22\#(2) \end{array}$$