Chapter 8.2, Problem 46E

To determine

To show: The changes that come under common difference while doubling each term in an arithmetic sequence.

Answer to Problem 46E

The common difference get double.

Explanation of Solution

Given information:

Doubling of each term in an arithmetic sequence.

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 : $a_n=a_1+(n-1)d$ .

Calculation:

Consider,

  $a_n=3n+1$

$n$	$1$	$2$	$3$	$4$	$5$	$6$
$a_n$	$\begin{array}{l}{a}_n=3(1)+1\\ =3+1\\ =4\end{array}$	$\begin{array}{l}{a}_n=3(2)+1\\ =6+1\\ =7\end{array}$	$\begin{array}{l}{a}_n=3(3)+1\\ =9+1\\ =10\end{array}$	$\begin{array}{l}{a}_n=3(4)+1\\ =12+1\\ =13\end{array}$	$\begin{array}{l}{a}_n=3(5)+1\\ =15+1\\ =16\end{array}$	$\begin{array}{l}{a}_n=3(6)+1\\ =18+1\\ =19\end{array}$

Therefore , the sequence is $4,7,10,13,16,19,\mathrm{........}$

  $\begin{array}{l}d=a_2-a_1=7-4=3\\ d=a_3-a_2=10-7=3\\ d=a_4-a_3=13-10=3\\ d=a_5-a_4=16-13=3\\ d=a_6-a_5=19-16=3\end{array}$

Now, doubling each term of the sequence we get,

  $8,14,20,26,32,38,\mathrm{.......}$

  $\begin{array}{l}d=a_2-a_1=14-8=6\\ d=a_3-a_2=20-14=6\\ d=a_4-a_3=26-20=6\\ d=a_5-a_4=32-26=6\\ d=a_6-a_5=38-32=6\end{array}$

We can conclude that the common difference has been double on doubling each term of the arithmetic sequence.