For x R , the least value of M for which a^ 1+x + a^ 1-x , M 2 , a^ 2x + a^ -2x (where a > 1 ) are three consecutive terms of an A.P. is given as 20 . The value of a is :

Step-by-step solution

Since the three terms are in A.P., we have: 2 ( M 2 ) = a^ 1+x + a^ 1-x + a^ 2x + a^ -2x M = a(a^x + a^ -x ) + (a^ 2x + a^ -2x ) Let y = a^x + a^ -x . For a > 1 and x R , the minimum value of y is 2 (which occurs at x = 0 ). We can express a^ 2x + a^ -2x in terms of y as: a^ 2x + a^ -2x = (a^x + a^ -x )^2 - 2 = y^2 - 2 Substituting this into the equation for M , we get: M = ay + y^2 - 2 Consider the function f(y) = y^2 + ay - 2 for y 2 . Since a > 1 , the vertex of this parabola is at y = - a 2 , which is negative.