The geometric sequence

In a geometric sequence each term is the term before it times the same number, $\displaystyle q$ – the common ratio:

• $\displaystyle 3, 6, 12, 24, \ldots$ is a geometric sequence with $\displaystyle {a_1}={3}$ and $\displaystyle {q}={2}$.
• $\displaystyle {\frac{a_{n+1}}{a_n}}={q}$ for every $\displaystyle n$: the quotient of any two neighbouring terms is the same number.
• In a geometric sequence $\displaystyle a_1\neq0$ and $\displaystyle q\neq0$, so no term is 0.

In a geometric sequence we multiply by the same number again and again, so it is worked with powers. The laws, for any whole exponents $\displaystyle m$, $\displaystyle n$:

The second law explains the exponent 0 and negative exponents:

• $\displaystyle {\frac{a^3}{a^3}}={a^0}$, and $\displaystyle {\frac{a^3}{a^3}}={1}$, so $\displaystyle {a^0}={1}$.
• $\displaystyle {\frac{a^2}{a^5}}={a^{-3}}$, and $\displaystyle {\frac{a^2}{a^5}}={\frac{1}{a^3}}$, so $\displaystyle {a^{-3}}={\frac{1}{a^3}}$.

From the first term to place $\displaystyle n$ we multiply by $\displaystyle q$ exactly $\displaystyle n-1$ times:

• In the sequence $\displaystyle 3, 6, 12, \ldots$: $\displaystyle {a_{10}}={3\cdot2^9}={1536}$.
• In the sequence $\displaystyle 48, 24, 12, \ldots$ ($\displaystyle {q}={\frac{1}{2}}$): $\displaystyle {a_n}={48\cdot\left(\frac{1}{2}\right)^{n-1}}={48\cdot2^{1-n}}$.

The ratio decides how the sequence behaves (when $\displaystyle a_1>0$):

• $\displaystyle q>1$: the sequence increases, faster and faster – $\displaystyle 3, 6, 12, 24, \ldots$
• $\displaystyle 0<q<1$: the sequence decreases, and the terms get close to 0 – $\displaystyle 48, 24, 12, 6, \ldots$
• $\displaystyle q<0$: the signs alternate – $\displaystyle 8, -4, 2, -1, \ldots$ ($\displaystyle {q}={-\frac{1}{2}}$), as in the graph:

In which place of the sequence $\displaystyle 3, 6, 12, \ldots$ is the term 384?

• $\displaystyle {3\cdot2^{n-1}}={384}$, that is $\displaystyle {2^{n-1}}={128}$.
• Try powers of 2: $\displaystyle {2^7}={128}$, so $\displaystyle {n-1}={7}$ and $\displaystyle {n}={8}$.

And which is the first term greater than 1,000?

• $\displaystyle 3\cdot2^{n-1}>1000$, that is $\displaystyle 2^{n-1}>333.3$.
• $\displaystyle {2^8}={256}$ is too small, and $\displaystyle {2^9}={512}$ is enough: $\displaystyle {n-1}={9}$, $\displaystyle {n}={10}$.
• $\displaystyle {a_{10}}={1536}$ is the first term greater than 1,000, and $\displaystyle {a_9}={768}$ is not yet.

Calculate.

$\displaystyle a_7$ in the sequence $\displaystyle 2, 6, 18, \ldots$

$\displaystyle a_7$ in the sequence $\displaystyle 4096, 1024, 256, \ldots$

Calculate with the laws of powers.

In which place of the sequence $\displaystyle 5, 10, 20, \ldots$ is the term 640?

The sequence: $\displaystyle {a_n}={3\cdot(-2)^n}$.
Work out $\displaystyle \frac{a_{n+1}}{a_n}$. What does the result prove?

In the sequence $\displaystyle \frac{1}{4}, \frac{3}{4}, \frac{9}{4}, \ldots$, which is the first term greater than 100? Find it by trial, and write its place.