# Whats the missing term in the geometric sequence 1, __, 49?

Jan 5, 2016

2^("nd") " term " ->7

$\textcolor{b l u e}{\text{Have a look at the way the sequence is built!}}$

#### Explanation:

As I understand it, a Geometric sequence is of structure:

$a {r}^{0} , a {r}^{1} , a {r}^{2} , a {r}^{3.} \ldots . .$ where r is the Geometric Ratio

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(brown)("Note that "r^0 = 1color(white)(...)" so "ar^0 = a

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In your case the first term $\to a {r}^{0} = 1$ so $a = 1$

Given that the third term $\to a {r}^{2} = 49$
as $a = 1 \text{ then } 1 \times {r}^{2} = 49$

So $\sqrt{{r}^{2}} = r = \sqrt{49} = 7$

The missing term $\to a {r}^{1} = 1 \times {7}^{1}$

${7}^{1} = 7 \textcolor{w h i t e}{\ldots}$so the missing term is 7