What is the 11th term of the sequence: \[2,8,32,128,...\]?

Hint: Let us consider, \[a\] be the initial term of the sequence and \[r\] be the common ratio then, the nth term of the sequence is,
\[{a_n} = a{r^{n - 1}}\].
If we see the given sequence the terms are increasing in a multiple of 4 by the last terms, that constant multiple is called the common ratio and the sequence is geometric progression.
Complete step-by-step solution:
It is given that; the sequence is \[2,8,32,128,...\]
We have to find the ${11^{th}}$term of the sequence: \[2,8,32,128,...\]
The given sequence is in geometric sequence.
So, the common ratio is \[\dfrac{8}{2} = \dfrac{{32}}{8} = 4\]
We know that, if \[a\] be the initial term of the sequence and \[r\] be the common ratio then, the nth term of the sequence is, \[{a_n} = a{r^{n - 1}}\]
We substitute, \[a = 2\] and \[r = 4\] we get,
\[{a_{11}} = 2 \times {4^{11 - 1}}\]
Simplifying we get,
\[{a_{11}} = 2097152\]
Hence, the ${11^{th}}$ term of the sequence \[2,8,32,128,...\]is \[2097152\].

Note: In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
The behaviour of a geometric sequence depends on the value of the common ratio.
If the common ratio is:
> positive, the terms will all be the same sign as the initial term.
> negative, the terms will alternate between positive and negative.
> greater than 1, there will be exponential growth towards positive or negative infinity (depending on the sign of the initial term).
> 1, the progression is a constant sequence.
> between −1 and 1 but not zero, there will be exponential decay towards zero (→ 0).
> −1, the absolute value of each term in the sequence is constant and terms alternate in sign.
less than −1, for the absolute values there is exponential growth towards (unsigned) infinity, due to the alternating sign.