Suppose a, b and c are in Arithmetic Progression and $$a^{2}, b^{2}$$ and $$c^{2}$$ are in Geometric Progression. If $$a<b<c$$ and a+b+c=$$\frac{3}{2}$$,, then the value of a=

Let us assume that the common difference of the A.P. is 'd'. 

Then, we can say that a = b - d, c = b + d

It is given that a + b + c = 3/2. i.e. b = 1/2. 

It is given that $$a^{2}, b^{2}$$ and $$c^{2}$$ are in Geometric Progression. Hence, we can say that 

$$b^4 = a^2*c^2$$

$$b^4 = (b - d)^2*(b + d)^2$$

$$b^4 = (b^2 - d^2)^2$$

$$\Rightarrow$$ $$(b^2+d^2-b^2)(b^2-d^2+b^2)=0$$ 

Therefore, $$(b^2 - d^2 + b^2) = 0$$. i.e. $$d = \dfrac{1}{\sqrt{2}}$$

Hence, a = b - d = $$\dfrac{1}{2}$$ - $$\dfrac{1}{\sqrt{2}}$$.