Question

# Which term of following GP is 256.$2$,$2\sqrt 2$,$4$…………..

Hint: In the questions we have to find the common ratio and then using the formula of the nth term of GP we can find which position the term 256 holds by equating it.

In the given series let's name the terms first i.e. ${a_1}$ = 2, ${a_2}$ =$2\sqrt 2$, ${a_3}$= 4
Let 256 be the ${a_n}$th term
Now, the common ratio $r$ is given by
$r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = .......... = \dfrac{{{a_n}}}{{{a_{n - 1}}}}$
∴ $r$=$\dfrac{{2\sqrt 2 }}{2}$=$\sqrt 2$
Using the formula of nth term
${a_n} = a{r^{n - 1}}$
$\Rightarrow 256 = 2{\left( {\sqrt 2 } \right)^{n - 1}}$
$\Rightarrow 128 = {\left( {\sqrt 2 } \right)^{n - 1}}$
Converting 128 in terms of power of 2
$\Rightarrow {2^7} = {2^{\dfrac{{n - 1}}{2}}}$
Since, bases are equal, therefore powers can also be equated.
$\Rightarrow 7 = \dfrac{{n - 1}}{2}$
$\Rightarrow 14 = n - 1$
$\therefore n = 15$
∴ 256 is the 15th term.

Note: Geometric progression is a sequence of numbers where each new term after the first is obtained by multiplying the preceding term by a constant r called common ratio. The common ratio is given by the formula $r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = .......... = \dfrac{{{a_n}}}{{{a_{n - 1}}}}$.