Question

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

Answer 1

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.