$$ \because $$$$\,\,\,$$a, b, c are in A.P. then

a + c = 2b

also it is given that,

a + b + c = $${{3 \over 4}}$$                          . . . .(1)

$$ \Rightarrow $$$$\,\,\,$$ 2b + b = $${{3 \over 4}}$$   $$ \Rightarrow $$$$\,\,\,$$ b = $${{1 \over 4}}$$  . . . . .(2)

Again it is given that, a², b², c² are in G.P. then

(b²)² = a²c²   $$ \Rightarrow $$  ac = $$ \pm $$ $${{1 \over {16}}}$$      . . . . (3)

From (1), (2) and (3), we get;

$$a \pm {1 \over {16a}}$$ = $${1 \over 2}$$  $$ \Rightarrow $$ 16a² $$-$$ 8a $$ \pm $$ 1 = 0

Case I : 16a² $$-$$ 8a + 1 = 0

$$ \Rightarrow $$$$\,\,\,$$a = $${1 \over 4}$$ (not possible as a < b)

Case II: 16a² $$-$$ 8a $$-$$ 1 = 0 

$$ \Rightarrow $$$$\,\,\,$$ a = $${{8 \pm \sqrt {128} } \over {32}}$$

$$ \Rightarrow $$$$\,\,\,$$ a = $${1 \over 4} \pm {1 \over {2\sqrt 2 }}$$

$$ \therefore $$$$\,\,\,$$ a = $${1 \over 4} - {1 \over {2\sqrt 2 }}$$   ($$ \because $$ a < b)