JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Sequences and Series

With common difference $d$ and first terms $A,A+1,A+2$, the chosen terms are $a=A+6d$, $b=A+1+8d$ and $c=A+2+16d$. In the determinant the second and third rows share the entries $7\to$ (second column) $17,17$ and third column $1,1$, so subtracting the third row from the second leaves only $2b-c$ in the first column, and the determinant equals $10(2b-c)$. The condition $10(2b-c)+70=0$ gives $c=2b+7$. Substituting, $A+2+16d=2(A+1+8d)+7$, which forces $A=-7$. Since $a=A+6d=29$, the common difference is $d=6$, and then $b=42$ and $c=91$. The new progression begins at $c-a-b=20$ with common difference $\frac{d}{12}=\frac{1}{2}$, so the sum of its first twenty terms is $\frac{20}{2}(2\cdot 20+19\cdot \frac{1}{2})=10(40+9.5)=495$.