Use the formula for the nth term of an arithmetic sequence

an = a1 + (n-1)d

with n=10  

a10 = a1 + (10-1)d

a10 = a1 + 9d

Substitute 75 for a10

75 = a1 + 9d

75 - a1 = 9d 

Since the right side is 9d, pick a number for a1
so that the left side will be a multiple of 9.  You don't
have to, but that's the way to avoid fractions.  If we pick
a1 to be 3, the left side will be 72 and that
is divisible by 9.

So let's pick a1 = 3

75 - 3 = 9d
    72 = 9d
     8 = d

So substitute a1 = 3 and d = 8 in

an = a1 + (n-1)d

an = 3 + (n-1)(8)

an = 3 + 8n-8

an = 8n - 5     <-- answer

Then the sequence goes:

3,11,19,27,35,43,51,59,67,75,83,91,99,107,...

Notice that the tenth term is 75.

You could give other answers by choosing different 
numbers for a1.

Edwin