Geometric Sequences with a Surd Ratio

Finding the ratio

A geometric sequence multiplies by the same number each time, the common ratio. Divide any term by the one before to find it.

In $\sqrt{2}$, 2, $2\sqrt{2}$, 4, … the ratio is $2 \div \sqrt{2} = \sqrt{2}$, and $2\sqrt{2} \div 2 = \sqrt{2}$ confirms it. A surd ratio works exactly like a whole-number one.

Every second term

$\sqrt{2} \times \sqrt{2} = 2$, so multiplying by $\sqrt{2}$ twice multiplies by 2. Every second term is twice the term two places before it: 4, then 8, then 16.

The terms in between carry the $\sqrt{2}$: $4\sqrt{2}$, then $8\sqrt{2}$.

The nth term

The nth term is $a \times r^{n-1}$. Here $a = \sqrt{2}$ and $r = \sqrt{2}$, so it is $\sqrt{2} \times \left(\sqrt{2}\right)^{n-1} = \left(\sqrt{2}\right)^{n}$.

An even power of $\sqrt{2}$ is a power of 2: $\left(\sqrt{2}\right)^{10} = \left(\left(\sqrt{2}\right)^{2}\right)^{5} = 2^{5} = 32$. An odd power leaves one $\sqrt{2}$ over: $\left(\sqrt{2}\right)^{7} = 2^{3} \times \sqrt{2} = 8\sqrt{2}$.

A missing term

If 1, x and 3 are consecutive terms, the ratio from 1 to x equals the ratio from x to 3: x / 1 = 3 / x. Multiplying both sides by x gives $x^{2} = 3$.

$x = \sqrt{3}$ gives 1, $\sqrt{3}$, 3, multiplying by $\sqrt{3}$ each time. $x = -\sqrt{3}$ works too: 1, $-\sqrt{3}$, 3 multiplies by $-\sqrt{3}$. A question says x > 0 when it wants only the positive one.

Ratios less than 1

The sequence 3, $\sqrt{3}$, 1, … divides by $\sqrt{3}$ each time, so its ratio is $\frac{1}{\sqrt{3}}$. Rationalized, that is $\frac{\sqrt{3}}{3}$: multiply top and bottom by $\sqrt{3}$.

The next term is $1 \times \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{3}$.

Adding the terms

Collect the whole numbers and the surds separately. $\sqrt{2} + 2 + 2\sqrt{2} + 4 = \left(2 + 4\right) + \left(\sqrt{2} + 2\sqrt{2}\right) = 6 + 3\sqrt{2}$.

The usual slips

Subtracting to find the ratio. $2 - \sqrt{2}$ is the gap between two terms; a geometric sequence divides.

Counting each $\sqrt{2}$ as a 2. $\left(\sqrt{2}\right)^{10}$ has ten factors of $\sqrt{2}$, which is only five factors of 2: 32, not $2^{10} = 1024$.

Squaring instead of rooting. $x^{2} = 3$ means $x = \sqrt{3}$, not 9.