If $$a, b, c, d$$   are four consecutive terms of an increasing A.P. then the roots of the equation $$\left( {x - a} \right)\left( {x - c} \right) + 2\left( {x - b} \right)\left( {x - d} \right) = 0$$        are

Solution :
If $$k\left( { > 0} \right)$$  be the common difference then the equation is
$$\eqalign{ & 3{x^2} - \left( {6a + 10k} \right)x + a\left( {a + 2k} \right) + 2\left( {a + k} \right)\left( {a + 3k} \right) = 0 \cr & \therefore \,\,D = {\left( {6a + 10k} \right)^2} - 4 \cdot 3 \cdot \left\{ {{a^2} + 2ak + 2{a^2} + 8ak + 6{k^2}} \right\} = 28{k^2} > 0. \cr} $$