Question

The $4^{th}$ term of a $G.P.$ is square of its second term, and the first term is $-3$. Determine its $7^{th}$ term.

Answer 1

Given,
$a_4=(a_2{)}^2$ and first term $=a=-3$

Step$1:$ Finding the value of common ratio.
We know, the $n^{th}$ term of $G.P.$ is
$a_n=a{r}^{n-1}$
So, $a_4=a{r}^{4-1}=a{r}^3$
and $a_2=ar$
According to the questions,
$a_4=(a_2{)}^2$
$\Rightarrow a{r}^3=(ar{)}^2$
$\Rightarrow a{r}^3=a^2{r}^2$
$\therefore r=a=-3$ (given)

Step$:2$ Finding $7^{th}$ term of a $G.P.$
$7^{th}$ term of $G.P.$ is
$a_7=a{r}^{7-1}$
$\Rightarrow a_7=a{r}^6$
$\Rightarrow a_7=(-3)(-3{)}^6$
$\Rightarrow a_7=(-3{)}^7$
$\therefore a_7=-2187$

Hence, $7^{th}$ term is $-2187$.