The sides of a `!ABC` are in A.P. such that a lt minimum (b,c). Then , cos A may be equal to

Options

A. `(3c-4b)/(2b)`
B. `(3c-4b)/(2c)`
C. `(4c-3b)/(2b)`
D. `(4c-3b)/(2c)`

Explanation

It is given that the sides of `Delta` ABC are in AP. And alt minimum (b, c}. Therefore, either b, c, a are in A.P. or c, b, a are in AP.
CASE I When b, c, a are in A.P. i.e. 2c = a + b
In this case we have
`cosA=(b^(2)+c^(2)-a^(2))/(2bc)=((b^(2)+c^(2))-(2c-b)^(2))/(2bc)=(4b-3c)/(2b)`
CASE II When c, b, a are in A.P. i.e. 2b = a + c
In this case, we have
`cosA=(b^(2)+c^(2)-a^(2))/(2bc)=(b^(2)+c^(2)-(2b-c)^(2))/(2bc)=(4c-3b)/(2c)`