How do I find the common ratio of a geometric sequence?

Question

A geometric sequence has its first term equal to $12$ and its fourth term equal to $-96$.

• How do I find the common ratio?
• And find the sum of the first $14$ terms

Answer 1

Here is a general method of such problems. Let $a_1,a_2,...$ be the GP series with common ratio $r$ then we must have $a_2=a_1r,a_3=a_2r=a_1r^2,...$ Now in your case we have, $a_1=12$ and $a_2=12r$ $a_3=12r^2$ and $a_4=12r^3=-96$ solving for $r$ in the last term we obtain $r=-2$.

Answer 2

Hint:

Let the first term of the sequence be $a_1$. Then $a_1 = 12$; the fourth term is $a_4 = -96$.

Since the sequence is geometric with ratio $r$, $a_2 = ra_1, a_3 = ra_2 = r^2 a_1,$ and so on.

With this fact, you can conclude a relation between $a_4$ and $a_1$ in terms of those two and $r$. With the former two known, you can solve for $r$.

From there, the formula for the sum of the first $n$ terms of a geometric sequence, with ratio $r$ and first term $a_1$, is given by

$$a_1 \cdot \frac{1-r^{n}}{1-r}$$

Answer 3

Terms of a geometric series are $a, ar, ar^2, ar^3, ...$,

where $a$ is the first term and $r$ is the common ratio.

In this case, $a=12$ and $ar^3=-96$, so $r^3=-8$, so $r=-2$.

The sum of the first $n$ terms of a geometric series (with $r\ne1$) is $a\dfrac{1-r^n}{1-r}$.

Can you take it from here?