Question

three terms of an arithmetic sequence are shown. f(1)=3, f(3)=6, f(7)=12 what recursive formula could define the sequence ? A. f(n+1)=2f(n) B. f(n+1)=f(n)+3 C. f(n+1)=f(n)+1.5 D. F(n+1)=0.5f(n)

Answer 1

$an\ arithmetic\ sequence:\\\\ d\ \ \rightarrow\ \ the\ difference\\\\f(n)=f(1)+(n-1)\cdot d\\\\f(n+1)=f(1)+n\cdot d\\\\f(n+1)-f(n)=f(1)+n\cdot d-[f(1)+n\cdot d- d]=\\\\=f(1)+nd-f(1) -nd+d=d\\\\\Rightarrow\ \ \ f(n+1)=f(n)+d$

$f(1)=3\ \ \ and\ \ \ f(3)=6\\\\f(3)=f(1)+(3-1)\cdot d\ \ \ \Rightarrow\ \ \ 6=3+2d\\\\ \ \ \Rightarrow\ \ \ 2d=3\ \ \ \Rightarrow\ \ \ d=1.5\\\\ f(n+1)=f(n)+d\ \ \ \Rightarrow\ \ \ f(n+1)=f(n)+1.5\\\\\\Ans.\ \ \ f(1)=3\ \ \ and\ \ \ f(n+1)=f(n)+1.5$