The sequence of terms u1, u2 , u3 ,… is defined by u(1) = 8, u(n+1) = au(n) + 1 where a is a negative constant. The third term of the sequence is 8. Find the value of a.

In an arithmetic series, the 8th term is twice the 3rd term and the 20th term is 110. Find the common difference.

2. 1) $u_1=8, \ u_{n+1}=au_n+1$
Then $u_2=au_1+1=8a+1$
$u_3=au_2+1=a(8a+1)+1=8a^2+a+1$
$u_3=8\Rightarrow 8a^2+a+1=0$ and you have to solve the quadratic

2) $\left\{\begin{array}{ll}a_8=2a_3\\a_{20}=110\end{a rray}\right.\Rightarrow\left\{\begin{array}{ll}a_1 +7r=2a_1+4r\\a_1+19r=110\end{array}\right.$
Now solve the system.