Marek, Jan, and Lucie solved the following task:
The sum of the first five terms of an increasing arithmetic sequence $(a_n)$ is equal to $10$. The terms $a_3$, $a_5$, and $a_{13}$, in the given order, form three consecutive terms of a geometric sequence. Find the formula for the $n$th term of the sequence $(a_n)$.
They all used the formula for the sum of the first five terms of an arithmetic sequence $(a_n)$ with the common difference $d$ and obtained: $$ \begin{gather} \frac{a_1+a_1+4d}{2}\cdot 5=10 \cr a_1+2d=2 \end{gather} $$
After that, each of them proceeded in a different way.
Marek reasoned that if $a_3$, $a_5$, and $a_{13}$ in the given order are three consecutive terms of a geometric sequence, then $$ a_1+4d=\frac{a_1+12d}{a_1+2d} $$ He substituted $2$ for $a_1+2d$ and calculated the common difference d of the sequence $(a_n)$: $$ \begin{gather} 2+2d=\frac{2+10d}{2} \cr d=\frac13 \end{gather} $$ Finally, he calculated $a_1$ and found the formula for the $n$th term of the sequence $(a_n)$: $$ \begin{gather} a_1=2-2d=\frac43 \cr a_n=\frac43+(n-1)\frac13 \cr a_n=1+\frac{n}3 \end{gather} $$
Jan reasoned this way: If $a_3$, $a_5$, and $a_{13}$ are three consecutive terms of a geometric sequence, then $$ \frac{a_1+4d}{a_1+2d}=\frac{a_1+12d}{a_1+4d} $$
He substituted $2$ for $a_1+2d$ and got: $$ \frac{2+2d}{2}=\frac{2+10d}{2+2d} $$
Then, he eliminated the fraction from the last equation and calculated the common difference: $$ \begin{gather} 4+8d+4d^2=4+20d \cr 4d(d-3)=0 \cr d=0,~d=3 \end{gather} $$
The last thing left for him to do was to determine $a_1$ and the formula for $a_n$.
For $d=0$ he got: $$ a_1=2,~a_n=2 $$ For $d=3$ he got $$ a_1=-4 $$ $$ a_n=-4+(n-1)\cdot 3=-7+3n $$
According to him, there are two arithmetic sequences of the given property.
Lucie remembered that for three consecutive terms of a geometric sequence $a_3$, $a_5$, $a_{13}$ must hold: $$ (a_1+4d)^2=(a_1+2d)(a_1+12d) $$ She removed parentheses on both sides and simplified the equation: $$ \begin{gather} a_1^2+8a_1 d+16d^2= a_1^2+14a_1 d+24d^2 \cr 6a_1 d+8d^2=0 \cr d(6a_1+8d)=0 \end{gather} $$ Then she reasoned if the last equation is to apply, then $d=0$ or $a_1=-\frac43 d$.
For $a_1=-\frac43 d$ she obtained $$ \begin{gather} -\frac43 d+2d=2 \cr d=3\end{gather} $$ From there $$a_1=-4$$ and $$a_n=-4+(n-1)\cdot 3=-7+3n$$
For $d=0$ she got $$a_1=2,~a_n=2$$
Lucie concluded that there are two arithmetic sequences of the given property.
Which one of them solved the example correctly?
None of them.
Marek
Jan and Lucie
Jan and Lucie would solve the example correctly if they excluded the solution $d=0$. In this case the sequence $(a_n)$ would be constant, which contradicts the assumption.