Question

Find the GP whose $4th$ and $7th$ terms are $\frac{1}{18}$ and $\frac{-1}{486}$ respectively.

Answer 1

$7_4=\frac{1}{18}$ $4_7=\frac{-1}{486}$

$a{r}^3=\frac{1}{18}$ $a{r}^6=\frac{-1}{486}$

$\Rightarrow \frac{a{r}^6}{a{r}^3}=\frac{-1/486}{1/18}\Rightarrow r^3=\frac{-1}{27}$ $r=\frac{-1}{3}$

$a(\frac{-1}{3}{)}^3=\frac{1}{18}\Rightarrow \frac{-a}{27}=\frac{1}{18}$

$a=\frac{-27}{18}=\frac{-3}{2}$

$GP=\frac{-3}{2},\frac{1}{2},\frac{-1}{6},\frac{1}{18},.......$