Tom and Ann solved the following task:
In a convergent infinite geometric sequence $(a_n )$ with positive terms, the first term is equal to $4$. Find the sum of all terms of this sequence if $a_3-a_5=\frac{32}{81}$.
Both Tom and Ann knew that in geometric sequence $(a_n )$ with the common ratio $r$ $$ a_3 =a_1 r^2,a_5 =a_1 r^4 $$ and so, they set up the equation: $$ a_1 r^2 -a_1 r^4=\frac{32}{81} $$ They substituted $4$ for $a_1$ and got: $$ \begin{gather} 4r^2-4r^4=\frac{32}{81} \cr 81r^2-81r^4=8 \end{gather} $$
Tom then continued as follows:
(1) He rewrote the last equation to: $$ 81(r^2 )^2-81r^2+8=0 $$
(2) Then he reduced the above equation to a quadratic equation in the variable $t=r^2$: $$ 81t^2-81t+8=0 $$
(3) He solved the quadratic equation using the quadratic formula: $$ \begin{gather} t_{1,2}=\frac{81\pm \sqrt{(-81)^2-4 \cdot 81 \cdot 8}}{2 \cdot 81} \cr t_{1,2}=\frac{81\pm \sqrt{81(81-32) }}{162} \cr t_{1,2}=\frac{81\pm {63}}{162} \cr t_1=\frac19,~ t_2=\frac89 \end{gather} $$
(4) Finally, he calculated the sums of infinite series $(a_n )$: $$ S_1=\frac{4}{1-\frac19}=\frac92,~~S_2=\frac{4}{1-\frac89}=36 $$ Tom concluded that there are two infinite geometric sequences with the given properties and their sums are: $$ S_1=\frac92,~S_2=36 $$
Ann proceeded the following way:
(1) She rewrote the last equation into the form: $$ (9r^2 )^2-9 \cdot (3r)^2+8=0 $$
(2) Then she used the substitution $$ u=3r $$ and changed the equation to: $$ u^4-9u^2+8=0 $$
(3) She solved the above equation by factoring: $$ \begin{gather} (u^2-1)(u^2-8)=0 \cr u^2=1 \mathrm{~or~} u^2=8 \cr u=\pm 1\mathrm{~or~} u=\pm \sqrt8 \end{gather} $$
(4) Finally, she determined the common ratio $r$ of the sequence: $$ r=\frac13, ~~r=\frac{\sqrt8}3 $$
(5) All that remained was to calculate the sums of the infinite series: $$ \begin{gather} S_1=\frac{4}{1-\frac13}=6 \cr S_2=\frac{4}{1-\frac{\sqrt8}{3}}=\frac{12}{3-\sqrt8}=12(3+\sqrt8)=36+24\sqrt2 \end{gather} $$
Ann concluded that there are two infinite geometric sequences with the given properties and their sums are: $$ S_1=6,~~S_2=36+24 \sqrt2 $$
The teacher asked their classmates to comment on their solutions. Which comment is right?
Ann got the correct result.
Tom got the correct result.
Ann's solution is not complete. Two results are missing: $r=-\frac13$ and $r=-\frac{\sqrt8}3$, which means that there are two other infinite geometric sequences with the given properties. Their sums are: $$ S_3=\frac{4}{1+ \frac13}=3 $$ and $$ S_4=\frac{4}{1+\frac{\sqrt8}{3}}=\frac{12}{3+\sqrt8}=12(3-\sqrt8)=36-24\sqrt2 $$
They both made a mistake right in the beginning. The $n$th term of the geometric sequence $(a_n )$ with common ratio $r$ is given by the formula $$ a_n=a_1 r^n $$ Therefore, they should have solved the equation: $$ a_1 r^3-a_1 r^5=\frac{32}{81} $$