Chapter 8.2, Problem 62E

a.

To determine

To find: The value of n.

Answer to Problem 62E

The value of $n=17$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{i=1}^n(3i+5)=544}$

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  $d=a_{n+1}-a_n;n\ge 1$ .

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

The sum of the first n terms of arithmetic series is

  $S_n=n\left(\frac{a_1+a_n}{2}\right).$

Calculation:

Consider ,

  ${\displaystyle \sum _{i=1}^n(3i+5)=544}$

Step1. Calculating the first and last terms

  $\begin{array}{l}{a}_i=3i+5\\ a_1=3(1)+5\\ =3+5\\ =8\\ a_n=3n+5\end{array}$

Step2. The sum is given.

  $\begin{array}{l}=S_n=544\\ =n\left(\frac{a_1+a_n}{2}\right)=544\\ =n\left(\frac{8+3n+5}{2}\right)=544\\ =n\left(\frac{3n+13}{2}\right)=544\\ =3n^2+13n=2\left(544\right)\\ =3n^2+13n-1088=0\end{array}$

  $\begin{array}{l}=n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =n=\frac{-13\pm \sqrt{169-4(3)(-1088)}}{2(3)}\\ =n=\frac{-13\pm \sqrt{169+13056}}{6}\\ =n=\frac{-13\pm \sqrt{13225}}{6}\\ =n=\frac{-13\pm 115}{6}\end{array}$

  $\begin{array}{l}=n=\frac{-13+115}{6}\\ =\frac{102}{6}\\ =17\end{array}$ or

  $\begin{array}{l}=n=\frac{-13-115}{6}\\ =-\frac{128}{6}\end{array}$

Therefore , $n=17$ .

Hence, the number of terms are $17.$

b.

To determine

To find: The value of n.

Answer to Problem 62E

The value of $n=23$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{i=1}^n(-4i-1)=-1127}$

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  $d=a_{n+1}-a_n;n\ge 1$ .

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

The sum of the first n terms of arithmetic series is

  $S_n=n\left(\frac{a_1+a_n}{2}\right).$

Calculation:

Consider ,

  ${\displaystyle \sum _{i=1}^n(-4i-1)=-1127}$

Step1. Calculating the first and last terms

  $\begin{array}{l}{a}_i=-4i-1\\ a_1=-4(1)-1\\ =-4-1\\ =-5\\ a_n=-4n-1\end{array}$

Step2. The sum is given .

  $\begin{array}{l}=n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =n=\frac{-(-6)\pm \sqrt{36-4(-4)(2254)}}{2(-4)}\\ =n=\frac{6\pm \sqrt{36+36064}}{(-8)}\\ =n=\frac{6\pm \sqrt{36100}}{(-8)}\\ =n=\frac{6\pm 190}{(-8)}\end{array}$

  $\begin{array}{l}=n=\frac{6+190}{(-8)}\\ =-\frac{196}{8}\end{array}$ or

  $\begin{array}{l}=n=\frac{6-190}{(-8)}\\ =\frac{-184}{-8}\\ =\frac{184}{8}\\ =23\end{array}$

Therefore , $n=23$ .

Hence, the number of terms are $23.$

c.

To determine

To find: The value of n.

Answer to Problem 62E

The value of $n=9$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{i=5}^n(7+12i)=455}$

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  $d=a_{n+1}-a_n;n\ge 1$ .

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

The sum of the first n terms of arithmetic series is

  $S_n=n\left(\frac{a_1+a_n}{2}\right).$

Calculation:

Consider ,

  ${\displaystyle \sum _{i=5}^n(7+12i)=455}$

The sum of the first 4 terms of arithmetic series is-

  ${\displaystyle \sum _{i=1}^4(7+12i)}$

Step1. The first and last terms

  $\begin{array}{l}{a}_i=7+12i\\ a_1=7+12(1)\\ =7+12\\ =19\\ a_4=7+12(4)\\ =7+48\\ \text{ = }55\end{array}$

Step2. The sum is −

  $\begin{array}{l}{S}_n=n\left(\frac{a_1+a_n}{2}\right)\\ S_4=4\left(\frac{19+55}{2}\right)\\ =4\left(\frac{74}{2}\right)\\ =148\end{array}$

The sum of the first 4 terms is 148.

Now,

Sum of first n terms − Sum of first 4 terms = ${\displaystyle \sum _{i=5}^n(7+12i)}$

Sum of first n terms − $148$ = $455$

Sum of first n terms = $455+148=603.$

Now, consider

  ${\displaystyle \sum _{i=1}^n(7+12i)=603}$

Step1. Calculating the first and last terms

  $\begin{array}{l}{a}_i=7+12i\\ a_1=7+12(1)\\ =7+12\\ =19\\ a_n=7+12n\end{array}$

Step2. The sum is given .

Now solving the equation we get,

  $\begin{array}{l}=n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =n=\frac{-(26)\pm \sqrt{676-4(12)(-1,206)}}{2(12)}\\ =n=\frac{-26\pm \sqrt{676+57,888}}{24}\\ =n=\frac{-26\pm \sqrt{588564}}{24}\\ =n=\frac{-26\pm 242}{24}\end{array}$

  $\begin{array}{l}=n=\frac{-26+242}{24}\\ =\frac{216}{24}\\ =9\end{array}$ or

  $\begin{array}{l}=n=\frac{-26-242}{24}\\ =\frac{-268}{24}\end{array}$

Therefore , $n=9$ .

Hence, the number of terms are $9.$

d.

To determine

To find: The value of n.

Answer to Problem 62E

The value of $n=15$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{i=3}^n(-3-4i)=-507}$

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  $d=a_{n+1}-a_n;n\ge 1$ .

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

The sum of the first n terms of arithmetic series is

  $S_n=n\left(\frac{a_1+a_n}{2}\right).$

Calculation:

Consider ,

  ${\displaystyle \sum _{i=3}^n(-3-4i)=-507}$

The sum of the first 2 terms of arithmetic series is-

  ${\displaystyle \sum _{i=1}^2(-3-4i)}$

Step1. The first and last terms

  $\begin{array}{l}{a}_i=-3-4i\\ a_1=-3-4(1)\\ =-3-4\\ =-7\\ a_2=-3-4(2)\\ =-3-8\\ \text{ =}-11\end{array}$

Step2. The sum is −

  $\begin{array}{l}{S}_n=n\left(\frac{a_1+a_n}{2}\right)\\ S_2=2\left(\frac{-7-11}{2}\right)\\ =2\left(-\frac{18}{2}\right)\\ =-18\end{array}$

The sum of the first 2 terms is $-18$ .

Now,

Sum of first n terms − Sum of first 2 terms = ${\displaystyle \sum _{i=1}^2(-3-4i)}$

Sum of first n terms − $(-18)$ = $-507$

Sum of first n terms = $-507-18=-525.$

Now, consider

  ${\displaystyle \sum _{i=1}^n(-3-4i)=-525}$

Step1. Calculating the first and last terms

  $\begin{array}{l}{a}_i=-3-4i\\ a_1=-3-4(1)\\ =-3-4\\ =-7\\ a_n=-3-4n\end{array}$

Step2. The sum is given .

Now solving the equation we get,

  $\begin{array}{l}=n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =n=\frac{-(-10)\pm \sqrt{100-4(-4)(1050)}}{2(-4)}\\ =n=\frac{10\pm \sqrt{100+16800}}{-8}\\ =n=\frac{10\pm \sqrt{16900}}{-8}\\ =n=\frac{10\pm 130}{-8}\end{array}$

  $\begin{array}{l}=n=\frac{10+130}{-8}\\ =\frac{140}{-8}\end{array}$ or

  $\begin{array}{l}=n=\frac{10-130}{-8}\\ =\frac{-120}{-8}\\ =15\end{array}$

Therefore , $n=15$ .

Hence, the number of terms are $15.$