$a_1$

Ania, Krysia and Ulla were to solve the following task:

The sum of the first three terms of an infinite geometric sequence is $14$, and the sum $S$ of all the terms of the sequence is $\frac{27}{2}$ . Calculate the first term $a_1$ of this sequence.

Ania solved the task this way: $$ \begin{gather} a_1+a_1 q+a_1 q^2=14 \cr \frac{a_1}{1-q}=\frac{27}{2} \end{gather} $$ Then she factored out $a_1$ in the first equation and isolated $a_1$ in the second equation: $$ \begin{gather} a_1 (1+q+q^2 )=14 \cr a_1=\frac{27}{2} (1-q) \end{gather} $$ Finally, she substituted $\frac{27}{2} (1-q)$ for $a_1$ in the first equation and solved the resulting equation: $$ \begin{gather} \frac{27}{2} (1-q)(1+q+q^2 )=14 \cr \frac{27}{2} (1+q^3 )=14 \cr 27(1+q^3 )=28 \cr q^3=\frac1{27} \cr q=\frac13 \end{gather} $$ So, she got: $$ a_1=\frac{27}{2} \left(1-\frac13\right)=9 $$

Krysia solved the task in the following way: $$ \begin{gather} a_1 (1+q+q^2 )=14 \cr \frac{a_1}{1-q}=\frac{27}2 \end{gather} $$ Then, in the first equation, she wrote the expression in parenthesis as a perfect square and isolated $a_1$ in the second equation: $$ \begin{gather} a_1 (1+q)^2=14 \cr a_1=\frac{27}2 (1-q) \end{gather} $$ Finally, she substituted $\frac{27}2 (1-q)$ for $a_1$ in the first equation and solved the resulting equation: $$ \begin{gather} \frac{27}{2} (1-q) (1+q)^2=14 \cr 27(1-q^3 )=28 \cr -27q^3=1 \cr q=-\frac13 \end{gather} $$ So, she got: $$ a_1=\frac{27}{2} \left(1+\frac13 \right)=18 $$

Ulla used the formulas for the sum of the first three terms of a geometric sequence and for the sum of an infinite convergent geometric series: $$ \begin{gather} a_1 \frac{1-q^3}{1-q}=14 \cr \frac{a_1}{1-q}=\frac{27}{2} \end{gather} $$ She then expressed $a_1$ from each equation: $$ \begin{gather} a_1= \frac{14(1-q)}{1-q^3} \cr a_1=\frac{27}{2} (1-q) \end{gather} $$ By comparing the expressions on the right sides, she got: $$ \frac{14(1-q)}{1-q^3}=\frac{27}{2} (1-q)
$$

Finally, she eliminated fractions from the above equation multiplying both sides by $\frac{2(1-q^3 )}{1-q}$ and determined $q$: $$ \begin{gather} 28=27-27q^3 \cr q=-\frac13 \end{gather} $$ All that remained was to calculate $a_1$: $$ a_1=\frac{27}{2} \left(1+\frac13 \right)=18 $$

Which one of them proceeded correctly in solving the task?