Proving that a sequence is geometric

A sequence is geometric when the ratio $\displaystyle \frac{a_{n+1}}{a_n}$ is the same number for every $\displaystyle n$, and no term is 0.
The proof below shows that the sequence $\displaystyle {b_n}={2\cdot3^{n-1}}$ is geometric:

• $\displaystyle {b_{n+1}}={2\cdot3^{n}}$.
• $\displaystyle {\frac{b_{n+1}}{b_n}}={\frac{2\cdot3^n}{2\cdot3^{n-1}}}={3^{n-(n-1)}}={3}$.
• The ratio is 3 for every $\displaystyle n$, so the sequence is geometric, and $\displaystyle {q}={3}$.

To show that a sequence is not geometric, one example is enough: two pairs of neighbouring terms with different ratios.
In the sequence $\displaystyle {a_n}={n\cdot2^n}$: $\displaystyle {a_1}={2}$, $\displaystyle {a_2}={8}$, $\displaystyle {a_3}={24}$.

• $\displaystyle {\frac{a_2}{a_1}}={4}$, but $\displaystyle {\frac{a_3}{a_2}}={3}$.
• The ratios are not equal, so the sequence is not geometric.

• The rule $\displaystyle {a_{n+1}}={5a_n}$, with $\displaystyle a_1\neq0$, says at once that the ratio is 5 for every $\displaystyle n$: the sequence is geometric.
• In the rule $\displaystyle {a_1}={1}$, $\displaystyle {a_{n+1}}={2a_n+1}$: $\displaystyle {a_2}={3}$, $\displaystyle {a_3}={7}$. $\displaystyle {\frac{a_2}{a_1}}={3}$ and $\displaystyle {\frac{a_3}{a_2}}={\frac{7}{3}}$, so the sequence is not geometric.

Here too $\displaystyle {a_n}={S_n-S_{n-1}}$ for every $\displaystyle n\geq2$, and $\displaystyle {a_1}={S_1}$.
In the example below $\displaystyle {S_n}={2^n-1}$ is given. We find $\displaystyle a_n$, and prove that the sequence is geometric:

• $\displaystyle {a_n}={(2^n-1)-(2^{n-1}-1)}={2^n-2^{n-1}}={2^{n-1}(2-1)}={2^{n-1}}$, for every $\displaystyle n\geq2$.
• $\displaystyle {a_1}={S_1}={2-1}={1}$, and also $\displaystyle {2^{0}}={1}$: the formula holds for every $\displaystyle n$.
• $\displaystyle {a_n}={2^{n-1}}$: a geometric sequence, $\displaystyle {a_1}={1}$ and $\displaystyle {q}={2}$.

The sequence: $\displaystyle {a_n}={5\cdot2^{n+1}}$.

Work out $\displaystyle \frac{a_{n+1}}{a_n}$.

Is the sequence geometric? What is its first term?

The sequence: $\displaystyle {a_n}={\frac{3}{2^{n}}}$.
Work out $\displaystyle \frac{a_{n+1}}{a_n}$, and explain why the sequence is geometric.

The sequence: $\displaystyle {a_n}={3^n+1}$.
Show by a counterexample that the sequence is not geometric.

The sum of the first $\displaystyle n$ terms of a sequence: $\displaystyle {S_n}={3^n-1}$.

Find $\displaystyle a_n$ for every $\displaystyle n\geq2$, and simplify.

Calculate $\displaystyle a_1$. Is the sequence geometric? If so, what is its ratio?

A sequence is defined for every natural $\displaystyle n$ by the recursion rule $\displaystyle {a_5}={8}$, $\displaystyle {n\cdot a_{n+1}}={4a_n-3}$.
From it: $\displaystyle {a_4}={\frac{35}{4}}$, $\displaystyle {a_5}={8}$, $\displaystyle {a_6}={\frac{29}{5}}$.
Is the sequence geometric? Explain.