Chapter 9, Problem 52RE

Solution

To identify: The sum of the first 10 terms of the arithmetic or geometric sequence. $94,91,88,\mathrm{...}$

The sum of the series $s_{10}=805$ .

Given information:

The given series $94,91,88,\mathrm{...}$ .

Explanation:

Consider the given sequence.

It is given that $94,91,88,\mathrm{...}$ .

Determine the value of common difference.

  $\begin{array}{c}r=a_2-a_1\\ r=91-94\\ r=-3\\ r=88-91\\ r=-3\end{array}$

This series is arithmetic series because the difference of consecutive term is constant.

Its mean that $a_1=94\text{and}r=91$ .

The sum nth term of an arithmetic sequence with first term $a_1$ and $a_7$ is given by: $s_n=n\left(\frac{a_1+a_n}{2}\right)$

Put the value of $a_1=94,n=10\text{and}r=-3$ into $a_n=a_1+\left(n+1\right)d$ .

  $\begin{array}{c}{a}_{10}=a_1+\left(n+1\right)d\\ a_{10}=94+\left(10-1\right)\left(-3\right)\\ a_{10}=94+\left(9\right)\left(-3\right)\\ a_{10}=94-27\\ a_{10}=67\end{array}$

Put the value of $a_1=94,n=10\text{and}r=-3$ into $s_n=n\left(\frac{a_1+a_n}{2}\right)$ .

  $\begin{array}{c}{s}_n=n\left(\frac{a_1+a_n}{2}\right)\\ s_{10}=10\left(\frac{94+67}{2}\right)\\ s_{10}=5\left(161\right)\\ s_{10}=805\end{array}$