Let a, b, c be in an AP and `a^2, b^2, c^2` be in GP. If a < b < c and `a + b+ c=3/2` then the value of a is

Options

A. `(1)/(2sqrt(2))`
B. `(1)/(2sqrt(3))`
C. `(1)/(2)-(1)/(sqrt(3))`
D. `(1)/(2)-(1)/(sqrt(2))`

Answer

D

Explanation

It is given that
a,b,c, are in A.P. `rArr" "2b=a+c`
`rArr" "2b=(3)/(2)-b" "[becausea+b+c=(3)/(2)("Given")]`
`rArr" "b=(1)/(2)rArra+c=1" "[becausea+c=2b]`
It is also given that
`a^(2),b^(2),c^(2)` are in G.P.
`rArr" "b^(2)=sqrt(a^(2)c^(2))rArrb^(2)=pmacrArrac=pm(1)/(4)" "[becauseb=(1)/(2)]`
CASE I When `ac=(1)/(2):` In this case, we have
`ac=(1)/(4)anda+c=1`
`:." "a+(1)/(4a)=1rArr(2a-1)^(2)=0rArra=(1)/(2)`
Thus, we have `a=c=b=(1)/(2)`, which is not possible as `altbltc` is given
CASE II When `ac=-(1)/(2):` In this case, we have
`ac=(1)/(4)anda+c=1`
`rArr" "a-(1)/(4a)=1`
`rArr" "4a^(2)-4a-1=0`
`rArr" "a=(4pmsqrt(16+16))/(8)=(1)/(2)pm(1)/(sqrt(2))`
If `a=(1)/(2)+(1)/(sqrt(2))," then "a+c=1rArrc=(1)/(2)-(1)/(sqrt(2))`
But, this not possible as `altc`. Thus, we have
`a=(1)/(2)-(1)/(sqrt(2))`