If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is

Let the first term be $$a$$ and common difference be $$d$$ .

Sum of first $$n$$ terms of an A.P. is given by  $$S_n = \dfrac{n}{2} \big[2a + (n-1)d \big] $$

Now, for the first 4 terms: $$ S_4 = \dfrac{4}{2}[2a + 3d] = 2(2a + 3d) = 6 \Rightarrow 2a + 3d = 3 \quad ...(1) $$

And, for first 6 terms: $$ S_6 = \dfrac{6}{2}[2a + 5d] = 3(2a + 5d) = 4 \Rightarrow 2a + 5d = \dfrac{4}{3} \quad ...(2) $$

Subtracting (1) from (2),
$$ (2a + 5d) - (2a + 3d) = \dfrac{4}{3} - 3 $$

$$ \Rightarrow 2d = \dfrac{4}{3} - \dfrac{9}{3} = -\dfrac{5}{3} \Rightarrow d = -\dfrac{5}{6} $$

Substituting this into (1):

$$ 2a + 3\left(-\dfrac{5}{6}\right) = 3 $$

$$ \Rightarrow 2a - \dfrac{5}{2} = 3 $$

$$ \Rightarrow a = \dfrac{11}{4} $$

Now, $$ S_{12} = \dfrac{12}{2}[2a + 11d] = 6[2a + 11d] $$

$$=6\ \cdot\left[\dfrac{11}{2}-\dfrac{55}{6}\right]=6\ \cdot\dfrac{33-55}{6}=-22$$