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
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 :
$$\frac{1}{2} - \frac{1}{{\sqrt 2 }}$$
Solution :
$$\eqalign{
& {\text{Given that }}a,b,c{\text{ are in A}}{\text{.P}}{\text{.}} \cr
& \Rightarrow \,\,{\text{2}}b = a + c \cr
& \Rightarrow \,\,{\text{but given }}a + b + c = \frac{3}{2} \cr
& \Rightarrow \,\,3b = \frac{3}{2} \cr
& \Rightarrow \,\,b = \frac{1}{2}{\text{ and then }}a + c = 1 \cr
& {\text{Again }}{a^2},{b^2},{c^2},{\text{ are in G}}{\text{.P}}{\text{.}} \cr
& \Rightarrow \,\,{b^4} = {a^2}{c^2} \cr
& \Rightarrow \,\,{b^2} = \pm ac \cr
& \Rightarrow \,\,ac = \frac{1}{4}{\text{ or }} - \frac{1}{4}\,\,{\text{and }}a + c = 1\,\,\,\,\,.....\left( 1 \right) \cr
& {\text{Considering }}a + c = 1{\text{ and }}ac = \frac{1}{4} \cr
& \Rightarrow \,\,{\left( {a - c} \right)^2} = 1 - 1 = 0 \cr
& \Rightarrow \,\,a = c{\text{ but }}a \ne c{\text{ as given that }}a < b < c \cr
& \therefore \,\,{\text{ We consider }}a + c = 1{\text{ and }}ac = - \frac{1}{4} \cr
& \Rightarrow \,\,{\left( {a - c} \right)^2} = 1 + 1 = 2 \cr
& \Rightarrow \,\,a - c = \pm \sqrt 2 \,\,{\text{but }}a < c \cr
& \Rightarrow \,\,a - c = - \sqrt 2 \,\,\,\,\,\,.....\left( 2 \right) \cr
& {\text{Solving }}\left( 1 \right){\text{ and }}\left( 2 \right){\text{ we get }}a = \frac{1}{2} - \frac{1}{{\sqrt 2 }} \cr} $$