Chapter 12.1, Problem 44E

To determine

To calculate: The sum of first 100 terms of the series $-5-4.8-4.6-\cdots$ .

Answer to Problem 44E

The sum of first 100 terms of the series is $490$ .

Explanation of Solution

Given information:

The series $-5-4.8-4.6-\cdots$ .

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 series $-5-4.8-4.6-\cdots$ .

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 $a_1,a_1+d,a_1+2d+\dots$ .

Compute the difference between the consecutive terms,

  $\begin{array}{c}{a}_2-a_1=-4.8+5\\ =0.2\end{array}$

And

  $\begin{array}{c}{a}_3-a_2=-4.6+4.8\\ =0.2\end{array}$

Therefore, common difference is $0.2$ .

According to question $a_1=-5$ and $n=100$ .

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

Apply it,

  $\begin{array}{c}{S}_n=\frac{100}{2}\left[2\left(-5\right)+\left(100-1\right)\left(0.2\right)\right]\\ =50\left[-10+99\left(0.2\right)\right]\\ =50\left(9.8\right)\\ =490\end{array}$

Thus, the sum of first 100 terms of the series is $490$ .