Chapter 9, Problem 57SGR

To determine

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

Answer to Problem 57SGR

The nth term of the series is: $a_n=2^{9-n}$

Explanation of Solution

Given:

The sequence: $256,128,64,\mathrm{......}$

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.

Calculation:

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 difference: $r=\frac{a_2}{a_1}=\frac{128}{256}=\frac{1}{2}$

Formula for nth term of GP series is: $a_n=a_1·r^{n-1}$

  $\text{Negative}\text{exponent}\text{rule}:{(\frac{1}{a})}^m=a^{-m}$

  $\text{Product}\text{rule}=a^m·a^n=a^{m+n}$

The nth term of the series: $a_n=256·{(\frac{1}{2})}^{n-1}=2^8·{(\frac{1}{2})}^{n-1}=2^8·{(2)}^{-(n-1)}=2^8·2^{-n+1}=2^{8-n+1}=2^{9-n}$

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

Thus, the nth term of the series is: $a_n=2^{9-n}$