SG. 1 The sum of 2 numbers is 20 , their product is 10 . If the sum of their reciprocals is $a$, find $a$. SG. $21^{2}-1=0 \times 2,2^{2}-1=1 \times 3,3^{2}-1=2 \times 4, \ldots, b^{2}-1=135 \times 137$. If $b>0$, find $b$. SG. 3 If the lines $x+2 y+1=0$ and $c x+3 y+1=0$ are perpendicular, find $c$.
SG. 4 The points $(2,-1),(0,1),(c, d)$ are collinear. Find $d$.
Reference: 1983 FG6.3, 1985 FSI.1, 1986 FSG. 1
Let the 2 numbers be $x$ and $y$.
$$
\begin{array}{l}
x+y=20 \text { and } x y=10 \\
a=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{x y}=2
\end{array}
$$

Reference: 1983 FI10.1, 1991 FI2.1
$$
\begin{array}{l}
135 \times 137=(136-1) \times(136+1)=136^{2}-1 \\
b=136
\end{array}
$$

Remark The original question is:
$$
1^{2}-1=0 \times 2,2^{2}-1=1 \times 3,3^{2}-1=2 \times 4, \ldots, b^{2}-1=135 \times 137 \text {, find } b \text {. }
$$
$b=136$ or -136 , there are 2 different answers!
Reference: 1983 FG9.3, 1985 FI4.1, 1986 FSG.2, 1987 FG10.2, 1988 FG8.2
$$
-\frac{1}{2} \times\left(-\frac{c}{3}\right)=-1 \Rightarrow c=-6
$$

Reference: 1984 FG7.3, 1986FG6.2, 1987 FG7.4, 1989 HG8
$$
\begin{array}{l}
\frac{d-1}{-6}=\frac{1-(-1)}{0-2} \\
d=7
\end{array}
$$