JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Sequences and Series

Writing the three terms as $\frac{A}{r},A,Ar$, their product is $A^3=27$, so $A=3$. The sum is $S=3\left(r+\frac{1}{r}+1\right)=3+3\left(r+\frac{1}{r}\right)$. For $r>0$, $r+\frac{1}{r}\ge 2$, giving $S\ge 9$; for $r<0$, $r+\frac{1}{r}\le -2$, giving $S\le -3$. Thus $S$ takes every value outside the open interval $(-3,9)$, so the excluded interval is $(a,b)=(-3,9)$. Therefore $a^2+b^2=9+81=90$.