Chapter 8, Problem 6RE

To determine

To find: an expression for the nth term of the given arithmetic sequence.

Answer to Problem 6RE

An expression for the nth term of the arithmetic sequence is $a_n=\frac{n+2}{n+1}$ .

Explanation of Solution

Given information:

An sequence is given as $\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5},\mathrm{...}$ .

Concept used:

An arithmetic sequence of n terms, has the form $a_1,a_2,a_3,\mathrm{...},a_n$ and should have a common difference between each two consecutive terms.

That is

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}$

Common difference can be defined by d .

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}=d$

nth term of the arithmetic sequence has the form

  $a_n=a_1+\left(n-1\right)d$

Where $a_1$ is the first term of the sequence, and d is the common difference.

Calculation:

Now, consider the given sequence.

  $\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5},\mathrm{...}=\frac{1+2}{1+1},\frac{2+2}{2+1},\frac{3+2}{3+1},\frac{4+2}{4+1}$

Both numerator and denominator are forming arithmetic sequences.

Consider dbe the common difference and $a_1$ as the first term of numerator.

  $a_1=3$

Common difference of denominator,

  $d=a_2-a_1=4-3=1$

So, the nth term of the arithmetic sequence can be found as

  $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ =3+\left(n-1\right)\left(1\right)\\ =3+n-1\\ =n+2\end{array}$

Consider d’be the common difference and ${a^{\prime }}_1$ as the first term of numerator.

  ${a^{\prime }}_1=2$

Common difference of denominator,

  $d^{\prime }={a^{\prime }}_2-{a^{\prime }}_1=3-2=1$

So, the nth term of the arithmetic sequence can be found as

  $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ =2+\left(n-1\right)\left(1\right)\\ =2+n-1\\ =n+1\end{array}$

So, the sequence will have nth term as

  $a_n=\frac{n+2}{n+1}$

Thus, the expression for the nth term of the arithmetic sequence is $a_n=\frac{n+2}{n+1}$ .