Question

Suppose a, b, c are in A.P. and a², b², c² are in G.P. If a < b < c and a + b + c =3/2, then the value of a is –

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

Answer 1

Answer: (D)

$\frac{1}{2} - \frac{1}{\sqrt{2}}$

Answer 2

Since a, b, c are in A.P.

\ b = a + d, c = a + 2d,

where d is a common difference, d > 0

Again since a², b², c² are in G.P.

\ a², (a + d)², (a + 2d)² are in G.P.

̃ (a + d)⁴ = a²(a + 2d)²

Or (a + d)² = ± a (a + 2d)

̃ a² + d² + 2ad = ± (a² + 2ad)

Taking (+) sign, d = 0 (not possible as a < b < c)

Taking (–) sign,

2a² + 4ad + d² = 0

̃ 2a² + 4a$\left( \frac{1}{2} - a \right) + \left( \frac{1}{2} - a \right)^{2}$= 0

$\left\lbrack \therefore a + b + c = \frac{3}{2} \Rightarrow a + d = \frac{1}{2} \right\rbrack$

̃ 4a² – 4a – 1 = 0

\ a = $\frac{1}{2}$± $\frac{1}{\sqrt{2}}$.

Here d =$\frac{1}{2}$– a > 0. So, a < $\frac{1}{2}$

Hence, a = $\frac{1}{2}$– $\frac{1}{\sqrt{2}}$.