Let a, b, c be in an AP and a^2, b^2, c^2 be in GP. If a < b < c and a + b + c = 3/2, then the value of a is

Let a, b, c be in an AP and a^2, b^2, c^2 be in GP. If a <… — JEE Main 2002 Mathematics PYQ

AIEEE 2002

Let

a, b, c

be in an AP and

a^2, b^2, c^2

be in GP. If

a < b < c

and

a + b + c = \frac{3}{2}

, then the value of

a

A.
$\frac{1}{2\sqrt{2}}$
B.
$\frac{1}{2\sqrt{3}}$
C.
$\frac{1}{2} - \frac{1}{\sqrt{3}}$
D.
$\frac{1}{2} - \frac{1}{\sqrt{2}}$

Show answer and solution

Correct option: D

Solution

Since a, b, c are in AP, b = 1/2 (so that a + b + c = 3/2 means a + c = 1, b = 1/2).

Since a², b², c² are in GP: b⁴ = a²c². So b² = ±ac, i.e., ac = ±1/4.

Since a < b < c with a + c = 1, if ac = 1/4 then a = c = 1/2, contradicting a < c. So ac = -1/4.

Now a and c are roots of t² - t - 1/4 = 0: t = (1 ± √2)/2.

Since a < b = 1/2 < c, we need a = (1 - √2)/2 = 1/2 - 1/√2.

[AIEEE 2002]

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