Question 831211
We are dealing here with an arithmetic progression where the the first term, a, is 58 and the common difference, d, is 15.

1.  The general formula for the nth term of an arithmetic progression is:

    a + (n - 1)d

   Substituting for a and d, we get 58 + 15(n - 1)

2.  When n = 10, we have 58 + 15 times 9 = 58 + 135 = 193

3.  The sum of the first n terms is given by the formula:
    (n/2)[2a + (n - 1)d]
    Substituting for a, n and d, we get:

    (10/2)[2 times 58 + 15 times 9] = 5[116 + 135] = 5 times 251 = 1255