JEE Main 2019 — Sequences And Series Question with Solution

Given, a, b, c are in G.P.

$$\therefore$$ b² = ac

In this equation ax² + 2bx + c = 0,

Discrimant, D = 4b² - 4ac

= 4ac - 4ac

= 0

Discrimant = 0 meand roots of the equation are equal.

Let both the roots of the equation = $$\alpha$$

$$\therefore$$ 2$$\alpha$$ = $$-\frac{2b}{a}$$

$$\Rightarrow$$ $$\alpha$$ = $$-\frac{b}{a}$$

As both the equations ax² + 2bx + c = 0 and dx² + 2ex + ƒ = 0 have a common root,

so $$-\frac{b}{a}$$ is also root of the equation dx² + 2ex + ƒ = 0.

$$\therefore$$ $$-\frac{b}{a}$$ satisfy the equation dx² + 2ex + ƒ = 0.

$$\therefore$$ $$d{\left(-\frac{b}{a}\right)}^2+2e\left(-\frac{b}{a}\right)+f=0$$

$$\Rightarrow$$ $$d{b}^2-2aeb+f{a}^2=0$$

$$\Rightarrow$$ $$dac-2aeb+f{a}^2=0$$

$$\Rightarrow$$ $$dc-2eb+fa=0$$

$$\Rightarrow$$ $$\frac{dc}{ac}-\frac{2eb}{ac}+\frac{fa}{ac}=0$$

$$\Rightarrow$$ $$\frac{dc}{ac}-\frac{2eb}{b^2}+\frac{fa}{ac}=0$$

$$\Rightarrow$$ $$\frac{d}{a}-\frac{2e}{b}+\frac{f}{c}=0$$

$$\Rightarrow$$ $$\frac{2e}{b}=\frac{d}{a}+\frac{f}{c}$$

$$\therefore$$ $$\frac{d}{a}$$, $$\frac{e}{b}$$, $$\frac{f}{c}$$ are in A.P.