(a) An arithmetic sequence $a_1, a_2, a_3, \ldots$ has common difference $d$, where $d<0$. The sum of the first $n$ terms of the sequence is denoted by $S_n$. Given that $\left|a_8\right|=\left|a_{13}\right|$, find the value of $n$ for which $S_n$ is maximum.
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(b) The terms $u_1, u_2$ and $u_3$ are three consecutive terms of a geometric progression. It is given that
$u_1, u_2$ and $u_3-32$
form an arithmetic progression, and that
$u_1, u_2-4$ and $u_3-32$
form another geometric progression. Find the possible values of $u_1, u_2$ and $u_3$.
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(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 $u_3=\frac{338}{9}$