Question

given two terms in a geometric sequence find the 8th term and the recursive formula . a4=-12 and a5=-6

Answer 1

A geometric sequence is defined by a starting point, $a$, and a common ratio $r$

The first term is $a$, and you get every next term by multiplying the previous one by r.

So, our terms are

$\left[\begin{array}{c|c}a_1&a\\a_2&ar\\a_3&ar^2\\a_4&ar^3=-12\\a_5&ar^4=-6\end{array}\right]$

We can see that when we pass from $a_4$ to $a_5$ the number gets halved ($-12 \mapsto -6$)

This implies that the common ratio is $r = \frac{1}{2}$

So, the table becomes

$\left[\begin{array}{c|c}a_1&a\\a_2&\frac{1}{2}a\\a_3&\frac{1}{4}a\\a_4&\frac{1}{8}a=-12\\a_5&\frac{1}{16}a=-6\end{array}\right]$

So, we can derive the starting point from either $a_4$ or $a_5$:

$\dfrac{1}{8}a = -12 \iff a = -12\cdot 8 = -96$

The sequence is thus

$\left[\begin{array}{c|c}a_1&-96\\a_2&-48\\a_3&-24\\a_4&-12\\a_5&-6\\a_6&-3\\\vdots&\vdots\end{array}\right]$

And the recursive formula is

$a_n = -\dfrac{96}{2^{n-1}}$