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Question: Let a,b,c (a \< b \< c) be in AP ; a<sup>2</sup>, b<sup>2</sup>, c<sup>2</sup> are in GP and a + b +...

Let a,b,c (a < b < c) be in AP ; a2, b2, c2 are in GP and a + b + c = 3/2 ; then a =

A

3+123\frac { \sqrt { 3 } + 1 } { 2 \sqrt { 3 } }

B

3−123\frac { \sqrt { 3 } - 1 } { 2 \sqrt { 3 } }

C

2−222\frac { \sqrt { 2 } - 2 } { 2 \sqrt { 2 } }

D

None

Answer

2−222\frac { \sqrt { 2 } - 2 } { 2 \sqrt { 2 } }

Explanation

Solution

Q a + b + c = 3b = 3/2 \ b =1/2

\ a = 12\frac { 1 } { 2 }– d & c = 12\frac { 1 } { 2 }+ d

Now (12\frac { 1 } { 2 }– d)2 , 14\frac { 1 } { 4 }, (12\frac { 1 } { 2 }+ d)2 are in GP

\ (using b2 = ac) ̃ 14\frac { 1 } { 4 }= d2 – 14\frac { 1 } { 4 } ̃ d = 1/ 2\sqrt { 2 }

\ a = 12\frac { 1 } { 2 }– d = 12\frac { 1 } { 2 }–12\frac { 1 } { \sqrt { 2 } }= 2−222\frac { \sqrt { 2 } - 2 } { 2 \sqrt { 2 } }