Chapter 12.1, Problem 57E

To determine

To calculate: Theamount of least and greatest prize awarded by radio station.

Answer to Problem 57E

Theleast prizeis $\$101$ and greatest prize is $\$1001$ .

Explanation of Solution

Given information:

The total of ten cash prizes is $\$5510$ . The difference between each successive prize is $\$100.$

Formula used:

A sequence with first term as $a_1$ and next term written as sum of preceding term and common difference then it is called an arithmetic sequence. It is denoted as $a_1,a_1+d,a_1+2d+\dots$ .

The sum of n terms of the arithmetic sequence is $S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ .

Calculation:

Consider the provided information that total of ten cash prizes is $\$5510$ . The difference between each successive prize is $\$100.$

Denote the least prize as first term of arithmetic sequence $a_1$ and greatest prize as last term $a_n$ .

Recall that a sequence with first term as $a_1$ and next term written as sum of preceding term and common difference then it is called an arithmetic sequence. It is denoted as $a_1,a_1+d,a_1+2d+\dots$ .

The sum of n terms of the arithmetic sequence is $S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ or $S_n=\frac{n}{2}\left[a_1+a_n\right]$

Here $S_n=5510,d=100\text{ and }n=10$ .

Apply it,

  $\begin{array}{c}{S}_n=\frac{n}{2}\left[a_1+a_n\right]\\ 5510=\frac{10}{2}\left[a_1+a_n\right]\\ 1102=a_1+a_n\\ a_1=1102-a_n\end{array}$

Again apply the sum of n terms of the arithmetic sequence $S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ ,

  $\begin{array}{c}5510=\frac{10}{2}\left[2\left(1102-a_n\right)+\left(10-1\right)100\right]\\ 1102=2204-2a_n+900\\ 2a_n=2002\\ a_n=1001\end{array}$

Since, $a_1=1102-a_n$ , so,

  $\begin{array}{c}{a}_1=1102-a_n\\ a_1=1102-1001\\ a_1=101\end{array}$

Thus, the least prize is $\$101$ and greatest prize is $\$1001$ .