Chapter 12.1, Problem 46E

To determine

To calculate: The value of $n$ for the arithmetic sequence for which sum of n terms is $-14$ , first term is $-7$ and common difference is $1.5$ .

Answer to Problem 46E

The value of $n$ for the arithmetic sequence is $8$ .

Explanation of Solution

Given information:

The arithmetic sequence has sum of n terms as $S_n=-14$ , first term as $a_1=-7$ and common difference as $d=1.5$ .

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 arithmetic sequence that has sum of n terms as $S_n=-14$ , first term as $a_1=-7$ and common difference as $d=1.5$ .

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 $S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ .

Apply it,

  $\begin{array}{c}-14=\frac{n}{2}\left[2\left(-7\right)+\left(n-1\right)\left(1.5\right)\right]\\ -28=n\left[-14+1.5n-1.5\right]\\ -28=-15.5n+1.5n^2\\ 15n^2-155n+280=0\end{array}$

Simplify it further as, apply the quadratic formula,

  $\begin{array}{c}3n^2-31n+56=0\\ n=\frac{31\pm \sqrt{{31}^2-4\left(3\right)\left(56\right)}}{6}\\ n=\frac{31\pm 17}{6}\end{array}$

That is, $n=\frac{31+17}{6}\text{ and }n=\frac{31-17}{6}$ .

Therefore, $n=8\text{ and }n=\frac{7}{3}$ .

Since n denotes the number of terms so it can take only integer values so reject $n=\frac{7}{3}$ .

Thus, the value of $n$ for the arithmetic sequence is $8$ .