First slide

Geometric progression

Question

Suppose a, b, c are in A.P. and $a^{2}, b^{2}, c^{2}$ , are in G.P . If $a < b < c and a + b + c = \frac{3}{2}$ then the value of a, is

Moderate

A

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

$\frac{1}{2 \sqrt{3}}$
C

$\frac{1}{2} - \frac{1}{\sqrt{3}}$
D

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

Solution

It is given that a, b, c are in A.P.
$\begin{array}{ll} \Rightarrow & 2 b = a + c \\ \Rightarrow & 2 b = \frac{3}{2} - b [∵ a + b + c = \frac{3}{2} (Given)] \\ \Rightarrow & b = \frac{1}{2} \Rightarrow a + c = 1 [∵ a + c = 2 b] \end{array}$
It is also given that $a^{2}, b^{2}, c^{2}$ are in G.P.
$\Rightarrow b^{2} = \sqrt{a^{2} c^{2}} \Rightarrow b^{2} = \pm ac \Rightarrow ac = \pm \frac{1}{4} [∵ b = \frac{1}{2}]$
CASE I: When ac =½ In this case, we have
$\begin{aligned} ac = \frac{1}{4} and a + c = 1 \\ ∴ & a + \frac{1}{4 a} = 1 \Rightarrow (2 a - 1)^{2} = 0 \Rightarrow a = \frac{1}{2} \end{aligned}$
Thus, we have a = c =b =½ which is not possible as a <b<c is given
CASE II: When ac = -½ In this case, we have
$\begin{aligned} ac = - \frac{1}{4} and a + c = 1 \\ \Rightarrow a - \frac{1}{4 a} = 1 \\ \Rightarrow 4 a^{2} - 4 a - 1 = 0 \\ \Rightarrow a = \frac{4 \pm \sqrt{16 + 16}}{8} = \frac{1}{2} \pm \frac{1}{\sqrt{2}} \\ If a = \frac{1}{2} + \frac{1}{\sqrt{2}}' then a + c = 1 \Rightarrow c = \frac{1}{2} - \frac{1}{\sqrt{2}} \end{aligned}$
But, this is not possible as a <c Thus, we have $a = \frac{1}{2} - \frac{1}{\sqrt{2}}$

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