JEE Main 2026 — 22 January, Morning Shift — Question 19

Sum of first 4 term $\mathrm{S}_{4}=6$ $\begin{gathered} \frac{4}{2}(2 a+3 d)=6 \Rightarrow 2 a+3 d=3 \end{gathered}$

Sum of first 6 terms $\mathrm{S}_{6}=4$ $\begin{gathered} \frac{6}{2}(2 a+5 d)=4 \Rightarrow 2 a+5 d=\frac{4}{3} \end{gathered}$

eq.(2) - eq. (1)

$(2 a+5 d)-(2 a+3 d)=\frac{4}{3}-3$

$\Rightarrow \mathrm{d}=-\frac{5}{6}$

$\therefore 2 a+3\left(-\frac{5}{6}\right)=3 \Rightarrow a=\frac{11}{4}$

$\mathrm{S}_{12}=\frac{12}{2}\left\{2 \times \frac{11}{4}+(12-1)\left(-\frac{5}{6}\right)\right\}$

$S_{12}=6\left(-\frac{22}{6}\right)=-22$