What are the first 4 terms of f(n-1)+15

What are the first 4 terms of f(n-1)+15

If f(1) = 7

$\begin{array}{rcll} f(2) &=& f(2-1)+15 \\ f(2) &=& f(1)+15 \qquad | \qquad f(1) = 7\\ f(2) &=& 7+15\\ f(2) &=& 22 \\\\ f(3) &=& f(3-1)+15 \\ f(3) &=& f(2)+15 \qquad | \qquad f(2) = 22\\ f(3) &=& 22+15\\ f(3) &=& 37 \\\\ f(4) &=& f(4-1)+15 \\ f(4) &=& f(3)+15 \qquad | \qquad f(3) = 37\\ f(4) &=& 37+15\\ f(4) &=& 52 \\\\ \cdots \end{array}$

arithmetic sequence: $\boxed{ ~ a_n=a_1+(n-1)\cdot d ~}$

$\begin{array}{rcll} f(n) &=& f(n-1)+15 \\ f(n)- f(n-1) &=& 15 \\ d &=& 15\\\\ \end{array}$

$\begin{array}{rcll} a_1 = f(1) &=& 7 \\ d &=& 15 \\\\ a_n &=& 7 + (n-1)\cdot 15 \\ a_n &=& 7 + 15n -15 \\ \mathbf{ a_n} & \mathbf{ = } & \mathbf{ -8 + 15n } \end{array}$

  If f(1) = 7, what are the first 4 terms of f(n-1)+15

7,  21, 35, 49, 63, 77...........