1.

Let ?, 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 isA. `(1)/(2sqrt2)`B. `(1)/(2sqrt3)`C. `(1)/(2) - (1)/(sqrt3)`D. `(1)/(2) - (1)/(sqrt2)`

Answer» Correct Answer - D
Since, a, b and c are in an AP
Let `a = A -D, b = A, c = A + D`
Given, `a + b + c = (3)/(2)`
`rArr (A -D) + A + (A +D) = (3)/(2)`
`rArr 3A = (3)/(2) rArr A = (1)/(2)`
`:.` The number are `(1)/(2) -D, (1)/(2), (1)/(2) + D`
Also, `((1)/(2) -D)^(2), (1)/(4), ((1)/(2) + D)^(2)` are in GP.
`:. ((1)/(4))^(2) = ((1)/(2) - D)^(2) ((1)/(2) + D)^(2) rArr (1)/(16) = ((1)/(4) - D^(2))^(2)`
`rArr (1)/(4) - D^(2) = +- (1)/(sqrt2)`
`:. a = (1)/(2) +- (1)/(sqrt2)`
so, out of the given values, `a = (1)/(2) - (1)/(sqrt2)` is the right choice


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