JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Sequences and Series

Write the three terms of the A.P. as $a-d,a,a+d$. Their sum is $3a=33$, so $a=11$. Their product is $a(a^2-d^2)=1155$, i.e. $11(121-d^2)=1155$, giving $121-d^2=105$, so $d^2=16$ and $d=\pm 4$. Taking $d=-4$, the first term is $15$ and the eleventh term is $15+10(-4)=-25$, which is the value offered among the choices.