Question

which term of the G.P $5,10,20,\mathrm{40....}$ is $1280$.

Answer 1

Consider the problem

In general, a geometric sequence is given by

$a,ar,a{r}^2,\cdots ,a{r}^{n-1}$

in the sequence $5,10,20,\mathrm{40....}$

First term is $a=5$

So, common ratio,

$\begin{array}{c}r=\frac{10}{5}\\ =2\end{array}$

Now, equate the term to found with the $n^{th}$ term

$\begin{array}{c}a{r}^{n-1}=1280\\ 5\times 2^{n-1}=1280\\ 2^{n-1}=\frac{1280}{5}\\ 2^{n-1}=256\\ 2^{n-1}=2^8\end{array}$

If the base are same then the powers must be equal

$\begin{array}{c}\Rightarrow n-1=8\\ \Rightarrow n=8+1\\ \Rightarrow n=9\end{array}$

Hence, $9^{th}$ of the sequence is $1280$