Question

Find the ${10}^{th}$ term and common ratio of the geometric sequence $\frac{1}{4},-\frac{1}{2},1,-2,....$

Answer 1

The given geometric progression is $\frac{1}{4},-\frac{1}{2},1,-2,.......$ where the first term is $a_1=\frac{1}{4}$, second term is $a_2=-\frac{1}{2}$ and so on.

We find the common ratio $r$ by dividing the second term by first term as shown below:

$r=\frac{-\frac{1}{2}}{\frac{1}{4}}=-\frac{1}{2}\times \frac{4}{1}=-2$

We know that the general term of an geometric progression with first term $a$ and common ratio $r$ is $T_n=a{r}^{n-1}$

To find the ${10}^{th}$ term of the G.P, substitute $n=10$, $a=\frac{1}{4}$ and $r=-2$ in $T_n=a{r}^{n-1}$ as follows:

$T_{10}=\frac{1}{4}(-2{)}^{10-1}=\frac{1}{4}(-2{)}^9=\frac{1}{4}\times (-512)=-\frac{512}{4}=-128$

Hence, the $10$th term of the given G.P is $-128$.