Chapter 7.2, Problem 56E

Solution

To calculate: The value of $n$ if ${\displaystyle \sum _{i=1}^n\left(10-3i\right)=-28}$ .

The value of $n=8$ .

Given information:

The given terms are ${\displaystyle \sum _{i=1}^n\left(10-3i\right)=-28}$ .

Formula used:

The sum nth term of an arithmetic sequence with first term $a_1$ and $a_n$ is given by: $s_n=n\left(\frac{a_1+a_1+\left(n-1\right)d}{2}\right)$

Calculation:

Consider the given sequence.

It is given that ${\displaystyle \sum _{i=1}^n\left(10-3i\right)=-28}$ .

Thus, $a_n=10-3n$

  $\begin{array}{c}{a}_n=10-3n\\ a_n=10-3n+3-3\\ a_n=7-\left(n-1\right)3\end{array}$

The nth term of an arithmetic sequence with first term $a_1$ and common difference $d$ is given by: $a_n=a_1+\left(n-1\right)d$ .

  $a_1=7\text{and}d=-3$

  $\begin{array}{c}{s}_n=n\left(\frac{a_1+a_1+\left(n-1\right)d}{2}\right)\\ -28=n\left(\frac{7+7+\left(n-1\right)\left(-3\right)}{2}\right)\\ -56=-3n^2+17n\end{array}$

Now solve this quadratic equation using the quadratic formula.

  $\begin{array}{c}n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ n=\frac{-17\pm \sqrt{{17}^2-4\cdot \left(-3\right)\cdot 56}}{2\cdot \left(-3\right)}\\ n=\frac{-17\pm \sqrt{961}}{-6}\\ n=\frac{-17\pm 31}{-6}\\ n=\frac{-17+31}{-6}\text{ or }n=\frac{-17-31}{-6}\\ n=-\frac{7}{3}\text{ or }n=8\end{array}$

Since the solution is positive and integer thus the only result is $8$ .