let

k = the number whose multiples we will discuss

the 7 consecutive multiples of k can be written as:

kn, k(n + 1), k(n + 2), k(n + 3), k(n + 4), k(n + 5), k(n + 6),

where n is integer

The average of 7 multiples of another number(not 5) is 28:

(1/7)(kn + k(n + 1) + k(n + 2) + k(n + 3) + k(n + 4) + k(n + 5) + k(n + 6)) = 28

(1/7)k(n + n + 1 + n + 2 + n + 3 + n + 4 + n + 5 + n + 6) = 28

(k/7)(7n + 6*7/2) = 28

k(n + 3) = 28

n + 3 = 28/k

The average of the first and third of these multiples is 20:

(kn + k(n + 2))/2 = 20

k(n + 1) = 20

n + 1 = 20/k

lets denote 1/k = t, now we have

n + 3 = 28t

n + 1= 20t

by solving the above system of equations we find:

n = 4

t = 1/4

click here to see the step by step solution of the system of equations:

1/k = t

1/k = 1/4

k = 4

the greatest of the multiples is k(n + 6) = 4(4 + 6) = 40.