Q57. - Algebra 1

Question

Write the equation for the $\mathit{n}\mathit{t}\mathit{h}$ term of each geometric sequence.

 $\mathbf{256}\mathbf{,}\mathbf{128}\mathbf{,}\mathbf{64}\mathbf{,}\mathbf{.}\mathbf{....}$

Step-by-Step Solution

Answer

The nth term of the series $256,128,64,\mathrm{.....}$ is ${\mathit{a}}_{\mathbf{n}}\mathbf{=}{\mathbf{2}}^{\mathbf{9}\mathbf{-}\mathbf{n}}$. 

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:  $256,128,64,\mathrm{.....}$

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 $a_1=256$ and the common ratio is:

 $\begin{array}{c}r=\frac{a_2}{a_1}\\ =\frac{128}{256}\\ =\frac{1}{2}\end{array}$.

Step 3. Find the n t h term.

In order to calculate the nth term substitute 256 for into the formula .

$\begin{array}{c}{a}_n=256\times {\left(\frac{1}{2}\right)}^{n-1}\\ =2^8\times 2^{-(n-1)}\\ =2^8\times 2^{-n+1}\\ =2^{8-n+1}\\ =2^{9-n}\end{array}$ 

The nth term of the series $256,128,64,\mathrm{.....}$ is $a_n=2^{9-n}$.