Sequence and Series – Arithmetic Progression – JEE Main 22 Jan 2026 Shift 1

Solution:

Let the first term be $a$ and the common difference be $d$.
Sum of first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.

Given $S_4 = 6$:
$$ \frac{4}{2}[2a + 3d] = 6 $$
$$ 2(2a + 3d) = 6 \Rightarrow 2a + 3d = 3 \quad \dots(1) $$

Given $S_6 = 4$:
$$ \frac{6}{2}[2a + 5d] = 4 $$
$$ 3(2a + 5d) = 4 \Rightarrow 2a + 5d = \frac{4}{3} \quad \dots(2) $$

Subtract equation (1) from (2):
$$ (2a + 5d) – (2a + 3d) = \frac{4}{3} – 3 $$
$$ 2d = \frac{4-9}{3} = -\frac{5}{3} $$
$$ d = -\frac{5}{6} $$

Substitute $d$ back into (1):
$$ 2a + 3\left(-\frac{5}{6}\right) = 3 $$
$$ 2a – \frac{5}{2} = 3 $$
$$ 2a = 3 + \frac{5}{2} = \frac{11}{2} \Rightarrow a = \frac{11}{4} $$

Now, find sum of first 12 terms ($S_{12}$):
$$ S_{12} = \frac{12}{2}[2a + (12-1)d] $$
$$ S_{12} = 6[2a + 11d] $$
$$ S_{12} = 6\left[ \frac{11}{2} + 11\left(-\frac{5}{6}\right) \right] $$
$$ S_{12} = 6 \left[ \frac{11}{2} – \frac{55}{6} \right] $$
$$ S_{12} = 6 \left[ \frac{33 – 55}{6} \right] $$
$$ S_{12} = 33 – 55 = -22 $$

Ans. (4)