Pavel, Patrick, and Ela had to solve this example: Determine all values of $m$ for which $\frac{m+1}{4}$, $\frac{m+3}{6}$, $\frac{m+9}{12}$ form three consecutive terms of an arithmetic sequence.
Pavel knew that if the numbers $a$, $c$, $b$ are consecutive terms of an arithmetic sequence, then $$ c=\frac{a+b}{2} $$ and so used the formula: $$ \frac{m+1}{4}+\frac{m+9}{12}=2\cdot \frac{m+3}{6} $$
Then he continued as follows: $$ \begin{gather} \frac{3m+3}{12}+\frac{m+9}{12}=\frac{m+3}{3} \cr \frac{4m+12}{12}=\frac{m+3}{3} \cr 4\frac{m+3}{3}=\frac{m+3}{3} \end{gather} $$ Finally, he simplified the equation by multiplying both sides by $\frac{3}{m+3}$ and got: $$ 4=1? $$ He did not forget the condition that $m\neq-3$. Pavel is now convinced that this example has no solution for $m\neq-3$ and that it has solution for $m=-3$. Really $$ -\frac{1}{2},\ 0,\ \frac{1}{2} $$ form three consecutive terms of an arithmetic sequence with common difference $d=\frac{1}{2}$.
Patrick remembered that an arithmetic sequence is a sequence of the form: $$ a,\ a+d,\ a+2d,\ldots $$ where $a$ is the first term and $d$ is the common difference of the sequence. He solved the example this way: $$ \begin{gather} \frac{m+9}{12}-\frac{m+3}{6}=2\left(\frac{m+3}{6}-\frac{m+1}{4}\right) \cr \frac{m+9-2\left(m+3\right)}{12}=2\cdot \frac{2\left(m+3\right)-3\left(m+1\right)}{12} \cr \frac{3-m}{12}=\frac{3-m}{6} \cr m=3 \end{gather} $$ As for him we get an arithmetic sequence if $m=3$. Really $$ \frac{4}{4},\ \frac{6}{6},\ \frac{12}{12} $$ are three consecutive terms of an arithmetic sequence with common difference $d=0$.
Ela thought that for the three consecutive terms of an arithmetic sequence must hold:
$$
\frac{m+3}{6}\div\frac{m+1}{4}=\frac{m+9}{12}\div\frac{m+3}{6}
$$
Then she continued:
$$
\begin{gather}
\frac{4(m+3)}{6(m+1)}=\frac{6(m+9)}{12(m+3)} \cr
48\left(m+3\right)^2=36(m+1)(m+9) \cr
4(m^2+6m+9)=\ 3(m^2+10m+9) \cr
m^2-\ 6m+9=0 \cr
\left(m-3\right)^2=0 \cr
m=3 \cr
\end{gather}
$$
Ela came to the conclusion that the given expressions do form three consecutive terms of an arithmetic sequence if $m=3$.
Which one of them proceeded correctly in solving?
None of them
Ela
Pavel
Patrick
Pavel’s way of solution was right, but he made a mistake in factoring out. $$ \frac{4m+12}{12}=\frac{m+3}{3} $$ It should have been: $$ \begin{gather} \frac{4\left(m+3\right)}{12}=\frac{m+3}{3}\cr \frac{m+3}{3}=\frac{m+3}{3} \end{gather} $$ From the above identity it is evident that the expressions $\frac{m+1}{4},\ \frac{m+3}{6},\ \frac{m+9}{12}$ form three consecutive terms of the arithmetic sequence for all $m\in\mathbb{R}$.