Question

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

Answer 1

We know that $a_n=a{r}^{n-1}$

where $a_n=n^{th}$ term of a G.P

$n$ is the number of terms.

$a$ is the first term.

$r$ is the common ratio.

Here, first term $=a=-3$

Also, $4^{th}$ term of G.P$a_4={\left(a_2\right)}^2$

Putting the above values in $a{r}^{4-1}={\left(a{r}^{2-1}\right)}^2$

$a{r}^3={\left(ar\right)}^2$

$a{r}^3=a^2{r}^2$

$\Rightarrow r=a$

Put $a=-3$ then $r=-3$

Thus, $r=-3,a=-3$

Using the formula, $a_n=a{r}^{n-1}$ let us find

$a_7=(-3){(-3)}^{7-1}=-3\times 729=-2187$

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