We can start by using the formula for the sum of the first n terms of an arithmetic sequence:

S_n = n/2[2a + (n-1)d]

where S_n is the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.

From the problem, we know that S_15 = 390 and the last term (a_15) is 47.

Using the formula, we can substitute the given values and solve for a and d:

390 = 15/2[2a + (15-1)d] 26 = 2a + 14d

We also know that a_15 = a + (15-1)d = 47.

Substituting this into the equation, we get:

47 = a + 14d

Now we have two equations with two variables:

26 = 2a + 14d 47 = a + 14d

We can solve for d by subtracting the second equation from the first:

26 - 47 = 2a - a + 14d - 14d -21 = a - 14d

Substituting this into the second equation, we get:

47 = -21 + 14d

Simplifying, we get:

d = 4

Substituting this into the equation a + 14d = 47, we get:

a + 14(4) = 47 a + 56 = 47 a = -9

Therefore, the first term is -9 and the common difference is 4.