Question

What is the $8$th term of the geometric sequence $3,9,27,......$ ?

Answer 1

Hint: Here we are finding the eighth term of the given geometry sequence by using the common ratio between the terms. And also using the formula for finding $n$th term of the geometric sequence.
Formula used:
Finding $n$th term of the geometric sequence using ${T_n} = a{r^{n - 1}}$
Where $a$ is the first term of the sequence and $r$ is the common ratio.
To calculate the common ratio of a geometric sequence , divide the second term of the sequence with the first term or simply find the ratio of any two consecutive terms by taking the previous term in the denominator, that is $r = \dfrac{{{a_1}}}{a}$ where ${a_1}$ is the second term of the sequence.

Complete step-by-step solution:
Given geometric sequence $3,9,27,.....$
First term of the sequence is $a = 3$ and the common ratio of the sequence is $r = \dfrac{{{a_1}}}{a} = \dfrac{9}{3} = 3$.
Now finding$n$th term of the geometric sequence using ${T_n} = a{r^{n - 1}}$
Substitute in the values of $a = 3$ and $r = 3$ we get,
${T_n} = \left( 3 \right){\left( 3 \right)^{n - 1}}$
Now we are going to find the $8$th term of the sequence ,
Substitute in the value of $n$ to find the $n$ the term, that is $n = 8$we get,
${T_8} = \left( 3 \right){\left( 3 \right)^{8 - 1}}$
${T_8} = \left( 3 \right){\left( 3 \right)^7}$
Using the property that is ${a^{m + n}} = {a^m}{a^n}$ , we get,
${T_8} = {\left( 3 \right)^8}$
Raise $3$ to the power of $8$ , we get,
${T_8} = 6561$
The eighth term of the geometric sequence is $6561$.

Note: Generally, to check whether a given sequence is geometric, one simply checks whether successive entries in the sequence all have the same ratio. The common ratio of a geometric sequence may be negative, resulting in an alternating sequence. We can also find the eighth term of the sequence by multiplying the previous term by the common ratio. but it takes too much effort to find the term.