If three numbers m, a, and k form an arithmetic sequence in that order, the sum of the numbers 21, and the product of the numbers is 315, what is the greatest number in the sequence

Let the common difference be d; 
 
then we have :-

m + m + d + m + 2d = 21    giving   3(m+d) = 21 and m +d = 7

m(m+d)(m+2d) = 315  giving  7m(m + 2d) = 315  and 7m(d + 7) = 315

m + d = 7
7md + 49m = 315

Substitute d = 7 - m in the second equation:-

7m(7 - m) + 49m = 315
49m -  7m^2 + 49m = 315
7m^2 - 98m + 315 =0
7 (m^2 - 14m + 45) = 0
7(m - 5)(m - 9) = 0

m = 5 or 9  and  d  = 7 - m = 2 or -2 

Try m = 5    then the sequence is  5 , 5+2 , 5+4 = 5 , 7 , 9
5 + 7 + 9 = 21  so m must be 5

and greatest number k  = 5 + 2(2) = 9 Answer