Question

Which term of the geometric sequence,
(i) $5,2,\frac{4}{5},\frac{8}{25},....$, is $\frac{128}{15625}$?
(ii) $1,2,4,8,....,$ is $1024$?

Answer 1

(i) The given geometric progression is $5,2,\frac{4}{5},\frac{8}{25},......$ where the first term is $a_1=5$, second term is $a_2=2$ and so on.

We find the common ratio $r$ by dividing the second term by first term as shown below:

$r=\frac{2}{5}$

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

Let the $n$th term of the given A.P be $T_n=\frac{128}{15625}$ and substitute $a=5$ and $r=\frac{2}{5}$ in $T_n=a{r}^{n-1}$ as follows:

$T_n=a{r}^{n-1}\Rightarrow \frac{128}{15625}=5{\left(\frac{2}{5}\right)}^{n-1}\Rightarrow \frac{128}{15625\times 5}={\left(\frac{2}{5}\right)}^{n-1}\Rightarrow \frac{2^7}{5^6\times 5}={\left(\frac{2}{5}\right)}^{n-1}\Rightarrow \frac{2^7}{5^{6+1}}={\left(\frac{2}{5}\right)}^{n-1}$

$\Rightarrow \frac{2^7}{5^7}={\left(\frac{2}{5}\right)}^{n-1}\Rightarrow {\left(\frac{2}{5}\right)}^7={\left(\frac{2}{5}\right)}^{n-1}\Rightarrow 7=n-1\Rightarrow n=7+1\Rightarrow n=8$

Hence, $8$th term of the given G.P is $\frac{128}{15625}$.

(ii) The given geometric progression is $1,2,4,8,,......$ where the first term is $a_1=1$, second term is $a_2=2$ and so on.

We find the common ratio $r$ by dividing the second term by first term as shown below:

$r=\frac{2}{1}=2$

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

Let the $n$th term of the given A.P be $T_n=1024$ and substitute $a=1$ and $r=2$ in $T_n=a{r}^{n-1}$ as follows:

$T_n=a{r}^{n-1}\Rightarrow 1024=1{\left(2\right)}^{n-1}\Rightarrow 2^{10}=2^{n-1}\Rightarrow 10=n-1\Rightarrow n=10+1\Rightarrow n=11$

Hence, $11$th term of the given G.P is $1024$.