If the sum of the two extreme terms of the AP is 16, then the sum of any two other terms, equally remoted from the end-terms is 16, too.


    It is WELL KNOWN property of any AP, the property which each student should learn when he or she studies the sum of an arithmetic progression.


Hence, in our case, if  = 16, then

        = 16,

        = 16.


Thus we can reformulate the original problem in this way:


    Find six numbers in AP, such that the sum of the two middle terms be 16 and the product of the two middle terms be 63.


So, the key moment is to find two number whose sum is 16 and the product is 63.


It is very elementary and standard problem. You can solve it using quadratic equation.

But it admits absolutely elementary MENTAL solution:  63 = 7*9  and 7 + 9 = 16.


Thus we found that  = 7  and   = 9.


Then the common difference is 9-7 = 2, and the AP is, obviously

    3, 5, 7, 9, 11, 13.


By the way, the reversed progression 

    13, 11, 9,  7,  5, 3

is the solution, too.