JEE Main 2019 — Sequences And Series Question with Solution

a, b, c are in G.P.

So, b = ar

and c = ar²

given   a + b + c = xb

$$\Rightarrow$$  a + br + ar² = x(ar)

$$\Rightarrow$$  1 + r + r² = xr

$$\Rightarrow$$  x = 1 + r + $$\frac{1}{r}$$

let sum of r + $$\frac{1}{r}$$ = M

$$\therefore$$  r² + 1 = Mr

$$\Rightarrow$$  r² $$-$$ Mr + 1 = 0

this quadratic equation will have

real solution when discriminant is $$\ge$$ 0

$$\therefore$$  b² $$-$$ 4ac $$\ge$$ 0

M² $$-$$ 4.1.1 $$\ge$$ 0

$$\Rightarrow$$  M² $$\ge$$ 4

M $$\ge$$ 2 or M $$\le$$ $$-$$ 2

$$\therefore$$  M $$\in$$ ($$-$$ $$\propto$$, $$-$$ 2] $$\cup$$ [2, $$\propto$$)

As   x = 1 + r + $$\frac{1}{r}$$

= 1 + M

$$\therefore$$  x $$\in$$ ($$-$$ $$\propto$$, $$-$$ 1] $$\cup$$ [3, $$\propto$$)

$$\therefore$$  x can't be 0, 1, 2.