Question

If the $4^{th}$ and $7^{th}$ terms of a G.P. are $54$ and $1458$ respectively, find the G.P.

Answer 1

It is given that the $4^{th}$ term of an G.P is $T_4=54$ and the $7$th term is $T_7=1458$

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

$T_4=a{r}^{4-1}\Rightarrow 54=a{r}^3.......\left(1\right)$

$T_7=a{r}^{7-1}\Rightarrow 1458=a{r}^6.......\left(1\right)$

Now divide equation 2 by equation 1 as follows:

$\frac{a{r}^6}{a{r}^3}=\frac{1458}{54}\Rightarrow r^3=27\Rightarrow r^3=3^3\Rightarrow r=3$

Substitute the value of $r$ in equation 1:

$a(3{)}^3=54$

$\Rightarrow 27a=54$

$\Rightarrow a=\frac{54}{27}$

$\Rightarrow a=2$

We also know the general term of G.P is $a,ar,a{r}^2,.....$, therefore, the terms of G.P are as follows:

$a=2$

$ar=2\times 3=6$

$a{r}^2=2(3{)}^2=2\times 9=18$ and so on.

Hence, the G.P is $2,6,18,.....$.