(a) $n=10$

(b) $u_1=2, u_2=10$ and $u_3=50$ or $u_1=\frac{2}{9}, u_2=\frac{26}{9}$ and $u_3=\frac{338}{9}$

(a)$\left|a_8\right|=\left|a_{13}\right|$
$\left|a_1+7 d\right|=\left|a_1+12 d\right|$
$a_1+7 d=a_1+12 d \quad$ or $\quad a_1+7 d=-a_1-12 a$
$7 d=12 d \quad 2 a_1+19 d=0$
$d=0 \quad a_1=-\frac{19}{2} d$
(rejected since $d<0$ )

Thus, $S_n=\frac{n}{2}\left(2 a_1+(n-1) d\right)=\frac{n}{2}(-19 d+n d-d)=\frac{d}{2} n(n-20)$ To find $n$ for max $S_n$ $S_n=\frac{d}{2} n(n-20)$ is a quadratic expression with negative coefficient of $n^2\left(\frac{d}{2}<0\right)$. When $S_n=0, n=0$ or $n=20$. Hence, $S_n$ is maximum at $n=\frac{0+20}{2}=10$.

(b)

Let $a$ and $r$ be the first term and common ratio of the geometric progression.
Then, $u_1=a, u_2=a r, u_3=a r^2$

$a r^2-2 a r+a=32$
$a(r-1)^2=32$

$(r-1)^2=\frac{32}{a}$ —–(1)

$\frac{a r^2-32}{a r-4}=\frac{a r-4}{a}$ —– (**)

$\left(a r^2-32\right) a =a^2 r^2-8 a r+16 $

$-32 a =-8 a r+16 \ a(8 r-32) =16 $

$a=\frac{2}{r-4}$. —– (2)

Substituting (2) into (1),
$r^2-2 r+1=16 r-64$
$r^2-18r+65=0$
$r = 5$ or $r=13$
$a = 2$ or $a=\frac{2}{9}$

Hence $u_1=2, u_2=10$ and $u_3=50$ or $u_1=\frac{2}{9}, u_2=\frac{26}{9}$ and $u_3=\frac{338}{9}$