JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Sequences and Series

Since $a,b,c$ are in G.P., $b^2=ac$. Look at the first quadratic $a{x}^2+2bx+c=0$: its discriminant is $4b^2-4ac=0$, so it has a repeated root $x=-\frac{b}{a}$. A common root means this value also satisfies $d{x}^2+2ex+f=0$, and substituting it and multiplying through by $a^2$ gives $d{b}^2-2aeb+f{a}^2=0$. Replacing $b^2$ with $ac$ and dividing by $ac$ gives $\frac{d}{a}-\frac{2e}{b}+\frac{f}{c}=0$, that is $\frac{d}{a}+\frac{f}{c}=2\cdot \frac{e}{b}$. This is exactly the condition for $\frac{d}{a},\frac{e}{b},\frac{f}{c}$ to be in A.P.