Recurrence relations: VJC 2024 Promo Q6

(a)(i)$u_1=p=2.4$

$u_2=12-\dfrac{24}{2.4}=2$

$u_3=12-\dfrac{24}{2}=0$

$u_4$ will be undefined and we are not able to generate an infinite sequence. Hence, $p$ cannot be 2.4.

6(a)(ii)If the sequence is constant, then all terms will be $p$.

$$p=12-\frac{24}{p}$$ $$p^2-12p+24=0$$ $$p=\frac{12\pm\sqrt{144-4(24)}}{2}=6\pm2\sqrt3$$ Since $p>5$, $p=6+2\sqrt3$.

6(b)$3v_{10}=5v_{19}$

$3(v_1+9d)=5(v_1+18d)$

$3v_1+27d=5v_1+90d$

$2v_1=-63d$

$$S_n=\frac n2\left(2v_1+(n-1)d\right)=\frac n2(-64d+nd)=\frac{nd}{2}(n-64)$$ Since $d<0$ and $S_n$ is a quadratic expression, $S_n$ is largest when $n=\dfrac{64}{2}=32$.

$$S_{32}=\frac{32d}{2}(32-64)=-512d$$ Alternative

$$S_n=\frac n2(-64d+nd)=\frac d2(n^2-64n)=\frac d2\left[(n-32)^2-1024\right]$$ Since $d<0$, $S_n$ is largest when $n=32$. $S_{32}=\dfrac{-1024d}{2}=-512d$.