$a_5$

Eve, Thomas and Rowan solved the task:

In an infinite geometric sequence $(a_n)$ the common ratio $q=\frac{\sqrt{3}}{3}$ and the sum of all its terms is equal to $51(\sqrt3+1)$. Calculate $a_5$.

All three used the formula for the sum of an infinite convergent geometric series with the first term $a_1$ and common ratio $q$: $$ S=\frac{a_1}{1-q} $$

Eve solved the task this way:

She expressed the term $a_5$: $$ a_5=a_1 q^4 $$ and substituted $S=51(\sqrt3+1)$ and $a_1=a_5 q^{-4}$ into the formula for the sum: $$ 51(\sqrt3+1)=\frac{a_5}{1-q} \cdot \frac{1}{q^4 } $$ From here she got: $$ a_5=51(\sqrt3+1)(q^4-q^5 ) $$ Finally, she substituted for $q$ and calculated $a_5$: $$ \begin{align} a_5 & =51(\sqrt3+1)\left(\frac19-\frac19 \cdot \frac{1}{\sqrt3} \right)\cr a_5 & =\frac{51}{9\sqrt3} (\sqrt3+1)(\sqrt3-1) \cr a_5 & =\frac{34}{3\sqrt3} \end{align} $$

Thomas presented this solution:

From the formula for the sum he expressed $a_1$: $$ a_1=S(1-q) $$ Since $S=51(\sqrt3+1)$ and $q=\frac{1}{\sqrt3}$, he continued: $$ \begin{align} a_1 &=51(\sqrt3+1) \left(1-\frac{1}{\sqrt3}\right) \cr \sqrt3 a_1&=51(\sqrt3+1)(\sqrt3-1) \cr a_1&=\frac{102}{\sqrt3} \end{align} $$ Finally, he used the formula $a_5=a_1+4q$ to calculate $a_5$: $$ \begin{align} a_5&=\frac{102}{\sqrt3}+\frac{4}{\sqrt3} \cr a_5&=\frac{106}{\sqrt3} \end{align} $$

Rowan substituted $51(\sqrt3+1)$ for $S$ and $\frac{\sqrt3}{3}$ for $q$ in the formula for the sum: $$ 51(\sqrt3+1)=\frac{a_1}{1-\frac{\sqrt3}{3}} $$ Then he continued: $$ \begin{align} 51(\sqrt3+1) & =\frac{3a_1}{3-\sqrt3} \cr 51(\sqrt3+1)(3-\sqrt3) & =3a_1 \cr 3a_1 & =51(3\sqrt3-\sqrt3) \cr a_1 & =34\sqrt3 \end{align} $$ Finally, he determined $a_5$: $$ \begin{align} a_5 & =a_1+4q \cr a_5 & =34{\sqrt3}+4 \cdot \frac{\sqrt3}{3} \cr a_5 & =\frac{102\sqrt3+4\sqrt3}{3} \cr a_5 & =\frac{106\sqrt3}{3} \end{align} $$ Which one of them solved the task correctly?