Chapter 9, Problem 51RE

Solution

To identify: The sum of the first 10 terms of the arithmetic or geometric sequence. $2187,729,243,\mathrm{...}$

The sum of the series $s_{10}=3280\overline{.4}$ .

Given information:

The given series $2187,729,243,\mathrm{...}$ .

Explanation:

Consider the given sequence.

It is given that $2187,729,243,\mathrm{...}$ .

Determine the value of common ratio.

  $\begin{array}{c}r=\frac{a_2}{a_1}\\ r=\frac{729}{2187}\\ r=\frac{1}{3}\\ r=\frac{243}{729}\\ r=\frac{1}{3}\end{array}$

This series is geometric series because the ratio of consecutive term is constant.

Its mean that $a_1=2187\text{and}r=\frac{1}{3}$ .

The sum nth term of a geometric sequence with first term and common ratio is given by: $s_n=\frac{a_1\left(1-r^n\right)}{1-r}$

Put the value of $a_1=2187,n=10\text{and}r=\frac{1}{3}$ into $s_n=\frac{a_1\left(1-r^n\right)}{1-r}$ .

  $\begin{array}{c}{s}_{10}=\frac{a_1\left(1-r^n\right)}{1-r}\\ s_{10}=2187\cdot \frac{1-{\left(\frac{1}{3}\right)}^{10}}{1-\left(\frac{1}{3}\right)}\\ s_{10}=2187\cdot \frac{1-\frac{1}{59049}}{\frac{2}{3}}\\ s_{10}=2187\cdot \frac{1}{59049}\cdot \frac{3}{2}\\ s_{10}=3280.\overline{4}\end{array}$