Question

Which term of the geometric sequence is 6,561 for , 1/27 , 1/9 , 1/3 , 1 ... ?

Answer 1

Step 1 - Write out the formula for the n-th term of a geometric sequence/progression.

`\begin{gathered} T_n=ar^(n-1) \\ \text{where, a is the first term of the sequence, and r is the common ratio.} \end{gathered}`

from the sequence question, a is 1/27

`\begin{gathered} To\text{ find the common ratio, r,} \\ Divide\text{ any two consecutive terms.} \\ \text{for instance, } \\ (T_2)/(T_1)=\frac{\text{Second term}}{\text{First term}}\text{ }=\text{ }((1)/(9))/((1)/(27))\text{ }=(1)/(9)\text{ x }(27)/(1)\text{ }=3 \end{gathered}`

Therefore the common ratio is r is 3 and the first term a is 1/27

We need both the first term and the common ratio to get the term that gives us a value of 6,561.

Step2 - substitute the values for the first term and common ratio into the n-th term formula earlier written.

`\begin{gathered} 6561=\text{ }(1)/(27)X3^(n-1) \\ 6561\text{ X 27 }=3^(n-1) \\ 177147\text{ }=3^(n-1) \\ 3^(11)\text{ }=3^(n-1) \\ \text{Comparing the powers on both sides of the equation} \\ 11\text{ }=\text{ n -1} \\ n\text{ }=\text{ 11}+1 \\ n\text{ }=\text{ 12.} \end{gathered}`

The term that gives 6561 in the sequence is 12