A sequence $u_1, u_2, u_3, \dots$ is such that $23u_{r+1} = 19u_r + 16$, for \(r

Question

A sequence $u_1, u_2, u_3, \dots$ is such that $23u_{r+1} = 19u_r + 16$, for $r > 0$ and $u_1 = 2$. Write down the value of $u_{10}$, giving your answer correct to $4$ decimal places.[1] It is given that as $r \to \infty$, $u_r \to l$. Show that $l = 4$.[1] Hence, find the smallest integer $r$ for which $u_r$ exceeds $99.9$% of $l$.[3] It is known that the $k$th term of this sequence is given by $u_k = 4 - 2\left(\frac{19}{23}\right)^{k-1}$. Given that a new series $S_n$ is defined by $S_n = \sum_{k=1}^{n} (u_k - c)$, where $c$ is a real constant. Express $S_n$ in the form of $An + B\left[1 - \left(\frac{19}{23}\right)^n\right]$, where $A$ is a constant in terms of $c$ and $B$ is a real number.[3] Determine the possible values of $c$ for $S_n$ to diverge as $n \to \infty$.[1]

Accepted Answer

(i) $3.6417$ (iii) $34$ (iv) $A=4-c$, $B=-\frac{23}{2}$ (v) ${c \neq 4}$

2025 SAJC P2 Q4

2025 SAJC P2 Q4