# How do you find the 13th term of the geometric sequence with a3 = 24 and a5 = 96?

Nov 24, 2015

${a}_{13} = 24576$

#### Explanation:

Well The definition of A geometric sequence with n elements;
a, ar,ar^2,ar^3, ar^4 ...ar^(n-1

r is a common constant with which every number is multiplied
Now

${a}_{5} = a \cdot {r}^{4} = 96 - - - - - - - - 1$

${a}_{3} = a \cdot {r}^{2} = 24 - - - - - - - - 2$

Now divide 1 by 2

$\frac{\cancel{a} \cdot {\cancel{r}}^{2} {r}^{2}}{\cancel{a} \cdot {\cancel{r}}^{2}} = \frac{96}{24} = 4$

${r}^{2} = 4 \implies r = 2$

Now come back to equation 2
$a \cdot {r}^{2} = 24$

Substitute
$a \cdot 4 = 24$

$\implies a = 6$

Now we have found the first term

Lets finish it

${a}_{13} = a \cdot {r}^{13 - 1} = 6 \cdot {2}^{6} \cdot {2}^{6} = 24576$

And were done