Anna, Tom, and Peter solved the following assignment:
The three consecutive terms of an arithmetic sequence $(a_n )$ are: $$x,~y-4~,y$$ The sum of these terms is equal to $6$. Calculate these terms of the sequence.
Tom used the formula for the sum of terms of an arithmetic sequence $$ S_n=\frac{n}{2} (a_1+a_n ) $$ where $n$ is the number of terms to be added, $a_1$ is the first term and $a_n$ is the last term to be summed. He also knew that if $x$, $y-4$, $y$ are three consecutive terms of an arithmetic sequence, then $$ y-4=\frac{x+y}{2} $$ and so, he solved the system of equations: $$ \begin{gather} 6=\frac{3}{2} (x+y) \cr 2y-8=x+y \end{gather} $$ He got the results: $x=2$ and $y=4$. The terms of the sequence are $$ 2,~0,~4 $$
Peter noticed that the third term of the given sequence is greater than the second term by $4$, which means that the common difference of the sequence $d=4$.
He expressed the first term of the sequence: $$ x=y-4-4 $$ and then used the formula for the sum of an arithmetic sequence $$ S_n=\frac{n}{2} (2a_1+(n-1)d) $$ where $n$ is the number of terms to be added, $a_1$ is the first term in the sequence and $d$ is the common difference.
He continued this way: $$ \begin{gather} 6=\frac{3}{2} (2(y-8)+2 \cdot 4) \cr y=6 \end{gather} $$ The terms of the sequence are: $$ -2,~2,~6 $$
Anna reasoned like this:
$$
\begin{gather}
x+y-4+y=6 \cr
x=10-2y
\end{gather}
$$
If the sequence were to be arithmetic, then it must hold:
$$
\frac{y-4}{x}=\frac{y}{y-4}
$$
She substituted $10-2y$ for $x$ in the above equation and eliminated the fractions:
$$y^2-8=(10-2y)y$$
Finally, she transformed the equation obtained in this way into the standard form of a quadratic equation and solved it:
$$
\begin{gather}
3y^2-10y-8=0 \cr
y_{1,2}=\frac{10\pm \sqrt{10^2-4 \cdot 3 \cdot (-8) }}{6} \cr
y_{1,2}=\frac{10\pm \sqrt{196}}{6} \cr
y_1=4 \mathrm{~and~} y_2=-\frac{2}{3}
\end{gather}
$$
For $y=4$, the terms of the sequence are $2,~0,~4$.
For $y=-\frac{2}{3}$, the terms of the sequence are $\frac{34}{3},~-\frac{14}{3},~-\frac{2}{3}$.
Which one of them proceeded correctly in solving?
Peter
Tom
Anna
None of them
The sum of the first $n$ terms of an arithmetic sequence can be calculated by one of the two formulas: $$ S_n=\frac{n}{2} (a_1+a_n ) $$ or $$ S_n=\frac{n}{2} (2a_1+(n-1)d) $$ where $n$ is the number of terms to be added, $a_1$ is the first term, $a_n$ is the last added term of the sequence and $d$ is the common difference. By substituting $$ a_n=a_1+(n-1)d $$ we get the second formula from the first one.