Question

Let $t_n$ represents the $n^{th}$ term of a G.P. If $t_3=2$ and $t_6=-\frac{1}{4},$ then the value of $t_{10}$ is

• A. $-\frac{1}{128}$
• B. $-\frac{1}{64}$
• C. $\frac{1}{128}$
• D. $\frac{1}{64}$

Answer 1

The correct option is B $-\frac{1}{64}$

Given : $t_3=2,t_6=-\frac{1}{4}$
Let the first term and common ratio of the G.P. be $a$ and $r$ respectively, then
$a{r}^2=2,a{r}^5=-\frac{1}{4}$
Dividing both, we get
$\Rightarrow r^3=-\frac{1}{8}\Rightarrow r=-\frac{1}{2}$
Now,
$a\times \frac{1}{4}=2\Rightarrow a=8$
Therefore, the ${10}^{th}$ term is
$t_{10}=a{r}^9\Rightarrow t_{10}=8{\left(\frac{-1}{2}\right)}^9=-\frac{1}{64}$