JEE Main 2018 — Sequences And Series Question with Solution

$$\because$$a, b, c are in A.P. then

a + c = 2b

also it is given that,

a + b + c = $$\frac{3}{4}$$                          . . . .(1)

$$\Rightarrow$$ 2b + b = $$\frac{3}{4}$$   $$\Rightarrow$$ b = $$\frac{1}{4}$$  . . . . .(2)

Again it is given that, a², b², c² are in G.P. then

(b²)² = a²c²   $$\Rightarrow$$  ac = $$\pm$$ $$\frac{1}{16}$$      . . . . (3)

From (1), (2) and (3), we get;

$$a\pm \frac{1}{16a}$$ = $$\frac{1}{2}$$  $$\Rightarrow$$ 16a² $$-$$ 8a $$\pm$$ 1 = 0

Case I : 16a² $$-$$ 8a + 1 = 0

$$\Rightarrow$$a = $$\frac{1}{4}$$ (not possible as a < b)

Case II: 16a² $$-$$ 8a $$-$$ 1 = 0 

$$\Rightarrow$$ a = $$\frac{8\pm \sqrt{128}}{32}$$

$$\Rightarrow$$ a = $$\frac{1}{4}\pm \frac{1}{2\sqrt{2}}$$

$$\therefore$$ a = $$\frac{1}{4}-\frac{1}{2\sqrt{2}}$$   ($$\because$$ a < b)