Geometric Sequences: Practice

The first term is the number written at the start of the list, the term in position $1$.

$a_1 = 4$

The value $3$ is the common ratio and $12$ is the second term, not the first.

The common ratio is any term divided by the term before it.

$r = \frac{12}{4} = 3$

Every ratio in the list is the same $3$, which is what makes the sequence geometric.

Each term is the one before it times the common ratio $r = 2$. Multiply the last term shown by $2$.

$80 \times 2 = 160$

Divide a term by the one before it. Because the list decreases, the ratio lies between $0$ and $1$.

$r = \frac{50}{100} = \frac{1}{2}$

For the 2nd term, $n = 2$, so the exponent is $n - 1 = 1$.

$a_2 = 3 \cdot 4^{1} = 12$

Adding the two numbers would give $7$, but a geometric sequence multiplies.

For the 3rd term, $n = 3$, so there are $n - 1 = 2$ factors of $r$.

$a_3 = 2 \cdot 3^{2} = 2 \cdot 9 = 18$

A ratio of $\tfrac{1}{3}$ means each term is one third of the one before it, so the list decreases.

$\frac{9}{27} = \frac{1}{3}$

Only $27, 9, 3, 1$ divides by $3$ each step; the others multiply by $3$ or by $2$.

The second term is the first term times the common ratio.

$a_2 = 5 \times 2 = 10$

The value $7$ comes from adding instead of multiplying.

Here $a_1 = 1$ and $r = 3$. The 5th term uses $n - 1 = 4$ factors of $r$.

$a_5 = 1 \cdot 3^{4} = 81$

Use $a_n = a_1 r^{n-1}$ with $n = 3$, so the exponent is $2$.

$a_3 = 1 \cdot 5^{2} = 25$

Count to position $2$: $3$ is first, $6$ is second.

$a_2 = 6$

Every term is built from where you start and the constant factor you multiply by, and nothing else.

$a_n = a_1 r^{n-1}$

This formula uses only $a_1$ and $r$, so those two numbers pin down the whole sequence.