Suppose a,b,c are in AP and a²,b²,c² are in GP. If a<b<c and a+b+c=3/2,then find the value of 'a'.

a=b=c=1/2? lol

By AM....... a+c =2b  ...so .....b =1/2 ....

and by GM.... b^2 = +ac .........so ac = +1/4

and a+c = 1..

.so a and c are roots of quadratic equation x^2 - x + 1/4....

and on solving a= 1/2 - 1/ 2^2

a=(1-2^1/2)/2

By AM....... a+c =2b  ...so .....b =1/2 ....

and by GM.... b^4 = (a^2.c^2 )^1/2 .........

so b^2 = + ac or  -ac ......

so ac = +1/4   or - 1/4

and a+c = 1..

.so a and c are roots of quadratic equation x^2 - x + 1/4.... or x^2 - x - 1/4

and on solving a= 1/2 - 1/ 2^2

Given,

       a, b, c are in A.P  => b = (a + c)/2

       and, given, that a + b + c = 3/2

                              => b + 2b = 3/2 => 3b = 3/2 =>  b = 1/2 -- (i)

Now, a + c = 2b = 2(1/2) => a + c = 1 -- (ii)

also, a² , b² , c² are in G.P => b² = ac = > ac = 1/4 -- (iii)

Now, (a - c)² = (a + c)² - 4 ac => (a - c)² = (1)² - 4(1/4) = 1 - 1 = 0 => (a - c)² = 0  => a - c = 0 -- (iv)

solving (ii) and (iii) we get ,

a = 1/2 , c = 1/2

Therefore, a = b = c = 1/2

hey hey rahul...u missed one point...its given dat a<b<c

Given, a + b + c = 3/2

since, a + c = 2b

=> b = 1/2

Now, a², b², c² are in G.P

=> b² = +ac 

=> ac = +b²

=> ac = 1/4 or ac = -1/4

but on taking ac = 1/4 we get, a = b = c = 1/2 [as i've got above]

but, on taking ac = -1/4 we get a step to the solution i.e., we get,

a - c = +√2 or a - c = -√2

But again, on taking a - c = +√2 we get  a > c, which can't be possible according to the question given.

Thus, as per the given condition a < b < c we are left with two eqns. only that is,

a + c = 1 -- (i)

a - c = - √2 -- (ii)

Now on solving these two eqns. we have :

a = (1 - √2) / 2 and c = (1 + √2) / 2

That is we have,

a = (1 - √2) / 2 , b = 1/2 , c = (1 + √2) / 2 , which also satisfies a < b < c