Show that a geometric sequence with first term 3 and seventh term 1/243 has two possible sums to infinity and find them.

Steps to Solve

1. Define the geometric sequence terms

In a geometric sequence, the $n$-th term can be expressed as $$ a_n = a \cdot r^{n-1} $$ where $a$ is the first term and $r$ is the common ratio.

Given that the first term, $a_1 = 3$, we have: $$ a = 3 $$

2. Express the seventh term

The seventh term is given as $$ a_7 = a \cdot r^{6} = \frac{1}{243} $$

Substituting the first term into this equation: $$ 3 \cdot r^{6} = \frac{1}{243} $$

3. Solve for $r^{6}$

To isolate $r^{6}$, divide both sides by 3: $$ r^{6} = \frac{1}{729} $$

This simplifies to: $$ r^{6} = (3^{-6}) $$ $$ \implies r = 3^{-1} \text{ or } r = -3^{-1} $$

This gives: $$ r = \frac{1}{3} \text{ or } r = -\frac{1}{3} $$

4. Calculate the sums to infinity

The sum to infinity of a geometric series is given by: $$ S = \frac{a}{1 - r} $$

For the first case when $r = \frac{1}{3}$: $$ S_1 = \frac{3}{1 - \frac{1}{3}} = \frac{3}{\frac{2}{3}} = \frac{3 \cdot 3}{2} = \frac{9}{2} = 4.5 $$

For the second case when $r = -\frac{1}{3}$: $$ S_2 = \frac{3}{1 - (-\frac{1}{3})} = \frac{3}{1 + \frac{1}{3}} = \frac{3}{\frac{4}{3}} = \frac{3 \cdot 3}{4} = \frac{9}{4} = 2.25 $$