Question

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

Answer 1

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