Q55. - Algebra 1

Question

Write the equation for the nth term of each geometric sequence.

$\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{.}\mathbf{....}$

Step-by-Step Solution

Answer

The nth term of the series $3,9,27,\mathrm{.....}$ is ${\mathit{a}}_{\mathbf{n}}\mathbf{=}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}$.

Step 1. State the concept used.

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. 

For example, the sequence $2, 6, 18, 54, ...$ is a geometric progression with common ratio 3.

The ration of $2^{nd}$ term and first term is called the common ratio.

Step 2. Find the common ratio.

The sequence: $-1,1,-1,1,\mathrm{.....}$

Each term in a geometric sequence can be expressed in terms of the first term $a_1$ and the common ratio $r$. Since each succeeding term is formulated from one or more previous terms, this is a recursive formula.

First-term $a_1=-1$ and the common ratio is:

 $\begin{array}{c}r=\frac{a_2}{a_1}\\ =\frac{1}{(-1)}\\ =-1\end{array}$

Step 3. Find the n t h term.

In order to calculate the nth term substitute $-1$ for $a_1$ into the formula$a_n=a_1{r}^{n-1}$.

$\begin{array}{c}{a}_n=-1\times {(-1)}^{n-1}\\ ={(-1)}^1\times {(-1)}^{n-1}\\ ={(-1)}^{1+n-1}\\ ={(-1)}^n\end{array}$

The $nth$ term of the series $-1,1,-1,1,\mathrm{.....}$ is $a_n={(-1)}^n$.